miimamimmKiBaiimamlisxa.^ 



o P% 



\«8* 



COMPLETE KEY 



UMMERE'S SURYEYmG; 



THE OPERATIONS OF ALL THE EXAMPLES NOT SOLVED IN 
THAT WORK ARE EXHIBITED AT LARGE. 



PRINCIPALLY DESIGNED 



TO FACILITATE THE LABOUR OF TEACHERS, 



AND TO ASSIST THOSE 



WHO HAVE NOT THE OPPOETUNITY OF THEIR INSTRUCTION. 

/ 
By SAMUEL ALSOP. 

ADAPTED. TO THE REVISED EDITION OF THE SURVEYING 
By ISAAC SHARPLESS, 

Author of "A Text-book of Geometry and Trigonombtky." 





PHILADELPHIA: 

PORTER & COATES. 



\ 



Copyright, 1884, 
BY PORTER & COATES. 






'<Pf7 



L^X)(r.C: 



KEY 



TO 



GUMMERE'S SURVEYING. 



PLANE TRIGONOMETRY. 



CASE 1. 

Example 3. (PI. 1, fig. 1.) 

Angle C=180°— A— B=46° 15'. 

As sin. A 19° 23' - - Ar. Co. 0.007499 

Is to sin. B 54° 22' - - - 9.909963 

So is EC 125 ■ 2.096910 



To AC 103.4 2.014372 

Again, 

As sin. A - Ar. Co. 0.007499 

Is to sin. C 46" 15' 9.858756 

SoisBC 2.096910 

To AB 91.87 1.968165 



Example 4. (PI. 1, fig. 2.) 
Angle C=90°— A=33° 12'. 

As sin. C 33° 1:2' Ar. Co. 0.261566 

Is to sin. B 90 - - - - 10.000000 

So is AB 53.66 1.729651 

To' AC 98 - - 1.991217 

As sin. C - - - ^ Ar. Co. 0.261566 

Is to sin. A 56° 48'^ 9.922603 

So is AB - - - 1.729651 

To BC 82 ----- 1.913820 



PLANE TRIGONOMETRY. [Case 2. 

Example 5. (PL 1, fig 2.) 
Angle C=90°— A=50° 50'. 

As sin. A 39° 10' - Ar. Co. 0.199573 

Ts to sin. B 90° - - 10.000000 

So is BC 407,37 2.609989 

To AC 645 2.809562 

As sin. A Ar. Co. 0.199573 

Is to sin. C 50° 50' - 9.889477 

SoisBC 2.609989 

To AB 500.1 - - -' 2.699039 



::;ase 2. 

Example 3. (PI. 1, fig. 1.) 

As AC 306 Ar. Co. 7.514278 

Is to AB 274 2.437751 

So is sin. B • 78° 13' 9.990750 

To sin. C - - 61 14 - 9-942779 

139 27 
180 

A - - 40 33 

As sin. B 78° 3' - Ar. Co. 0.009515 

Is to sin. A 4113^ 23' -9.812988 

So is AC 306 2.485721 

To BC 203.4 2.308224 

Example 4, (PI l,fig. 2.) 

As AC 272 - - - - Ar. Co. 7.565431 

•Is to AB 232 2.365488 

So is sin. B - 90° lOOOOMO 

To sin. C . - 58° 32' - 9.930919 

A - - 31° 28' 



/ 



Cases.] PLANE TRIGONOMETRY 



As sin. B Ar. Co. 0.000000 

Is to sin. A 31° 28' 9.717673 

So is AC - - - 2.434569 

ToiBC 142 2.152242 



Example 5. (PI. 1, fig. 2.) 

As AC 150 Ar. Co. 7.823909 

Is to BC 69 - - 1.838849 

So is sin. B - 90° 10.000000 



To sin. A - 27° 23' 9.662758 

C - 62° 37' 

As sin. B Ar. Co. 0.000000 

Is to sin. C 62° 37' 9.948388 

So is AC 2.176091 

ToAB 133.2 - - 2.124479 



CASE 3. 

Example 2. (PL 1, fig. 3.) 

A + C^180°-B^3^. 15' 
2 2 

As AB+BC 185 Ar. Co. 7.732828 

IstoAB— BC33 - - 1.518514 

So is tang. -^±^ - - 39° 15' - - - - - 9.912240 

° 2 

To tang. 5ll^ - - 8° 18' 9.163582 

^ 2 = 

C - 47° 33' 



As- sin. A 30° 57' ------ - Ar. Co. 0.288792 

Is to sin. B 101° 30' 9.991193 

So is BC 76 - - - - 1.880814 

To AC 144.8 2.160799 



PLANE TRIGONOMETRY. [Case 4. 



Example 3. (PL 1, fig. 2.) 

:45° 



C+A_180°— B_._^ 



a 2 

As AB+BC 1677 Ar. Co. 6.775467 

Is to AB— BC 103 2.012837 

So is tariff. ?±A - - 45° 10.000000 

^ 2 

To tang. ^^ - - 3° 31' 8.788304 

C - - 48° 31' ~ 

As sin. A 41° 29' Ar. Co. 0.178878 

Is to sin. B 90 10.000000 

So is BC 787 2.895975 

To AC 1188 3.074853 



Case 4.] 



RIGHT-ANGLED TRIANGLES, 







CASE 4. 








Rule 2. 






Example 2. (Pi. : 


flg. 4.) 


AC 


47 






AB 


64 


Ar. Co. 


8.193820 


BC 


34 
2)145 


Ar. Co. 


8.468521 


Half sum 


72.5 


- -. - - 


- 1.860838 


Difference 


25.5 


- - - - 


■ - 1.406540 




2)19.929219 


( 


:os. i B 

B 

EXAMP 


22° 49' - - 


. - 9.964609 




45° 38' 






LE 3. (PI. 1, 


fig. 4.) 


AB 


108 






BC 


54 


Ar. Co. 


8.267606 


AC 


88 
2)250 


Ai. Co. 


8.055517 




IS"! 




- - 2 096910 




17 




- . 1.230449 










2)19.650482 


Co 


s.iC 
C 


48° 2' - 


=. . 9.825241 




96° 4' 











RIGHT ANGLED TRIANGLES. 

First Method. 

Example 2. (PL 1, fig. 5.) 
Making AC radius, CB is sine of A, and AB is cos, A ; hence, 



8 PLANE TRIGONOMETRY. [Case 4. 

As radius Ar. Co. 0.000000 

Is to sin. A 27° 46' 9.668267 

So is AC 36.57 1.563125 

ToBC 17.04 - -' 1-231392 

And, 

As radius - - - - Ar. Co. 0.000000 

Is to COS. A 9.946871 

So is AC 1.563125 

To AB 32.36 - - - 1.509996 

Example 3. (PI. 1, fig. 5.) 

Making AC radius, we have CB the sine, and AB the cosine ot 
A; hence, 

As sin. A 42° 9' Ar. Co. 0.173230 

Is to radius 10.000000 

So is BC 193.6 2.286905 

To AC 288.5 2.46013 5 

And, 

As sin. A Ar. Co. 0.173230 

Is to cos. A - 9.870047 

So is BC - - - - 2.286905 

To AB 213.9 2.330182 

Example 4. (PI. 1, fig. 6.) 

Making the base AB radius, we have BC the tangent of A ; making 
AC the radius, we have AB the cosine of A ; hence, 

As AB 46.72 Ar. Co. 8.330497 

Is to BC 57.9 1.762679 

So is rad. 10.000000 

To tang. A 51° 6' ----- 10.093176 

And', 

As COS. A Ar. Co. 0.202066 

Is to rad. 10.000000 

SoisAB - - 1.669503 

To AC 74.4 1.871569 

/ 



HEIGHTS AND DISTANCES. 



Second Method. — By Logarithms. 
Example 3. 



Hypothenuse 
Base - - - 


403 
321 




Sum - - - - 


724 - - 


. log. 2.859739 


Difference - - 


82 - - 


- -" 1.913814 




2)4.773553 


Perpendicular 


243.65 - 


- - - 2.386776 



Example 4. 
Perpendicular 27.2 - - - log. 1.434569 

Base - 



- 31.04 - - 


2.869138 
- . 1.491922 1.491922 


23.835 - - 


- - 1.377216 


54.875 


1.739374 


3 41.27 - - 


2)3.231296 
1.615648 







APPLICATION OF PLANE TRIGONOMETRY TO THE 
MENSURATION OF DISTANCES AND HEIGHTS. 

Example 1. (See fig. 54, Surveying.) 

Angle C=180°— A— B=56° 23'. 

To find AC: 

As sin. C 56° 23' Ar. Co. 0.079480 

Is" to sin. B 49° 23' - 9.880289 

So is AB 500 yards 2.698970 

To AC 455.8 --.... 2.658739 



10 PLANE TRIGONOMETRY. 

To find BC : 

As sin. C Ar. Co. 0.079480 

Is to sin. A 74° 14' - . . . 9.983345 

So is AB - - - - 2.698970 

To BC 577.8 2.761795 



Example 2. (Fig. 55, Surveying.) 

As BC + AC 1575 Ar. Co. 6.802719 

~ Is to BC— AC 105 - - - 2.021189 

A+B 

So is tang. — ^ 62° 10' 10.277379 

To tang. -^^=?- 7° 12' 9.101287 

B - - 54° 58' 

As sin. B Ar. Co. 0.086818 

Is to sin. C 55° 40' 9.916859 

So is AC 735 2.866287 

To AB 741.2 2.869959 



Example 3. (Fig. 56, Surveying.) 
Angle CAD=180— ADC— ACD=31° 10' 

To find AC : 

As sin. CAD 31° 10' ----- - Ar. Co. 0.286065 

Is to sin. ADC 53° 30' 9.905179 

So is CD 300 - - - ' 2.477121 

To AC 465.98 ----- 2.668365 

Angle CBD= 180— BCD— BDC = 22° 55' 

To find CB: 

As sin. CBD 22° 55' Ar. Co. 0.409613 

Is to sin. CDB 98° 45' 9.994916 

So is CD - 2.477121 

To CB 761.47 •. . . . 2.881650 



HEIGHTS AND DISTANCES. 11 

To find AB: 

As BC + AC 1227.45 Ar. Co. 6.910995 

Is to BC— AC 295.49 2.470542 

So is tariff, —-?±^^ - 71° 30' - - - - 10.475480 
^ 2 

* To tang. ^AB— CBA . 350 44' ... . 9.857017 
^ 2 = 

CBA - 35° 46 



As sin CBA ...-.---- Ar. Co. 0.233226 

Is to sin. BCA 37° . - . 9.779463 

So is CA 465.98 - - - 2.668367 

To AB 479.8 ----- 2.681056 



Example 4. (Pig. 57, Surveying.) 

To find C : 

AB 3 

AC 2 Ar. Co. 9.698970 

BC 1.8 - ... - Ar. Co. 9.744727 

2)6.8 

Half sum 3.4 0.531479 

Diflference .4 -. —1.602060 

2)19.577236 

Cos. iC' - 52° 4' ----- - 9.788618 

C - 104° 8' 

To find BD: 

As -sin. D 17° 47' Ar. Co. 0.515105 

Is to sin. C 104° 8' ^ 9.986651 

So is BC 1.8 0.255273 



ToBD 5.715 - - - 0.757029 



12 PLANE TRIGONOMETRY. 

To find CD : 

As sin. D „ . . Ar. Co 0.515105 

Is to sin. DBC 58° 5' - - 9.928815 

So is BC - - - ■ 0.255273 

To CD 5,003 - - 0.699193 



Example 5. (Fig. 58, Surveying,) 

To find BAC: 

BC 7.2 

AB 12 - - . - Ar. Co. 8.920819 

AC 8 ~ - - - Ar. Co. 9.096910 

27.2 

Half sum 13.6 . = ..... 1.133539 

Difference 6.4 .-.--.. 0.806180 

2)19.957448 

Cos. h BAC 17° 47'i .... 9.978724 

BAC 35° 35 

To find AE: 

As sin. AEB 136° ...... Ar. Co. 0.158229 

Is to sin. EBA 19° ........ . 9.512642 

So is AB 12 .... - 1.079181 

To AE 5.624 -....--.... 0.750052 

To find ACE: 

As AC + AE 13.624 ..... Ar. Co. 8.865696 
Is to AC— AE 2.376 - - - 0.375846 

So is tang. ^^^ + ^^^ . 84° 42'i - . - 11.033291 
To tang. AEC— ACE _ g^ ij . „ . 10.274833 

AEC . 146 44 

ACE - 22° 41' 

/ 



HEIGHTS AND DISTANCES. 

To find AD: 

As sin. ADC 19° Ar. Co. 0.487358 

Is to sin. ACD 22= 41' 9.586179 

So is AC 8 0.903090 

- To AD 9.476 0.976627 

To find CD: 

As sin. ADC Ar. Co. 0.487358 

Is to sin. DAC 138° 19' ....... - 9.822830 

So is AC - - - . 0.903090 

To DC 16.34 1.213278 

To find BD : 

As sin. ADB 44° ....... Ar. Co. 0.158229 

Is to sin. BAD 102° 44' -..-..- - 9.989186 

So is AB 12 - - - 1.079181 

To DB 16.85 1.226596 



13 



Example 6. (Fig. 59, Surveying.) 



To find ABC : 








AC 46 








AB 50 




Ar.Co. 


8.301030 


BC 40 




Ar. Co. 


8.397940 


2)136 








Half sum "' 68 


- 


- . - . 


- 1.832509 


Difference 22 


30° 


8' - - 


. 1.342423 




2)19.873902 


Cos. ^ ABC ^ 


- 9.936951 


ABC 1 


60° 


16' 





14 PLANE TEIGONOMETRY. 

To find CD and CE : 

As sin. ADC 60° 16' Ar. Co. 0.061309 

Is to radius 10.000000 

So is AC 46 .. - 1.662758 

To CD - - 52.98 1.724067 

CE=iCD = 26.49 

Also CAE=90°— ADC=90°— ABC=29° 44'. 

Example 7. (Fig. 61, Surveying.) 
Making DE radius, EC is tangent of D ; hence, 

As radius Ar. Co. 0.000000 

Is to tang. D 47° 30' 10.037948 

So is DE 100 2.000000 

To EC - 109.13 2.037948 

EB - 5 :^= 

EC . 114.13 



Example 8. (Fig. 62, Surveying.) 

To find DC: 

As sin. ACD 25° Ar. Co. 0.374052 

Is to sin. CAD 26° 30' ----■... . 9.649527 

So is AD 75 ft. -----...- . 1.875061 

To DC 79.18 ft. 1.898640 

To find BC: 

As radius Ar. Co. 0.000000 

Is to sin. CDB 51° 30' . 9.893544 

So is CD - 1.898640 

To CB 61.97 ft. ..-..-... . 1.792184 

Example 9. (Fig. 63, Surveying.) 
To find DC: 

As sin. ACD 23° 50' . ... Ar. Co. 0.398535 

Is to sin. CAD 44° 9.841771 

So is AD 134 - - - - . - 2.127105 

To DC 230.4 - - - 2.362411 

/ 



HEIGHTS AND DISTANCES. 15 

To find CE: 

As sin. CED 141° Ar. Co. 0.201128 

Is to sin. CDE 16° 50' 9.461782 

So is CD - - 2.362411 

To CE 106 2.025321 

To find EB: 

As radius Ar. Co. 0.000000 

Is to sin. CDB 67° 50' 9.966653 

So is CD 2.362411 

To CB - - 213.3 - - .- 2.329064 

CE - - 106 = 

BE - - 107.3 



Example 10. (Fig. 64, Surveying.) 

To find BD: 

As sin. CDB 17° 15' ..... Ar. Co. 0.527914 

Is to sin. BCD 23° 45' 9.605032 

So is CB 60 - - - 1-'^7^1^1 

To BD 81.49 1-911097 

To find AD: 

As BD + BA 121.49 ..... Ar. Co. 7.915460 

Is to BD— BA 41.49 - - - - - - . . 1.617943 

So is tang. BAD+BDA _ ggo ^q> ... io.427262 

^ 2 

To tang. BAD— BDA _ ^^o 35' . - - 9.960665 
BDA - 27° 5' 

As sin. ADB 27° 5' Ar. Co. 0.341716 

Is to sin. ABD 41° .----.-- - 9.816943 

So is AB 40 1.602060 

To AD 57.64 1.760719 



16 PLANE TRIGONOMETRY. 



Example 11. (Fig. 65, Surveying.) 

To find AD: 

As sin. CAD 27° Ar. Co. 0.342953 

Is to sin. ACD 138° - - - 9.825511 

So is CD 132 ----- - 2.120574 

To AD 2.289038 

To find AB; 

As sin. ABD 109° Ar. Co. 0.024330 

Is to sin. ADB 8° 9.143555 

So is AD 2.289038 

To AB 28.64 1.456923 



PRACTICAL QUESTIONS. 17 



PRACTICAL QUESTIONS. 

Example 1. (PI. 1, fig. 2.) 
Making AB radius, BC is tangent of A. 

As radius - - - - Ar. Co. 0.000000 

Is to tang. A 52° 30' - - - 10.115020 

So is AB 85 - ... 1 1.929419 

To BC 110.8 2.044439 

Example 2. (PI. 1, fig. 2.) 

Make AB radius, then will BC be the tangent of A ; making AC 
radius, AB will be the cosine of A ; hence, 

As radius - - - Ar. Co. 0.000000 

Is to tang. A 61° 45' - - - 10.269767 

So is AB 73 - 1.863323 

To BC 135.9 - - - 2.133090 

And, 

As cosine A .- Ar. Co. 0.324845 

Is to radius ----.:..------- 10.000000 

So is AB 1.863323 

To AC 154.2 2.188168 



Example 3. (PI. 1, fig. 7.) 

To find BD. We have in the triangle ABD, the angles and side 
AB. Hence, 

As sin. ADB 31° Ar. Co. 0.288161 

Is to sin. BAD 100° ^ 9.993351 

So is AB 339 2.530200 

To BD 648.2 - - - 2.811712 

2 



18 PLANE TRIGONOMETRY. 

Again, in ABC we have the angles, and side AB, to find BC. Thus, 

As sin. ACB 22° 30' Ar. Co. 0.417160 

Is to sin. BAC 36° 30' 9.774388 

So is AB 339 - : 2.530200 

To BC 526.9 2.721748 

In DBC we have the sides DB and BC, and included angle DBC= 
72°. To find the side DC. Thus 

BCD + BDC = 1 80°— 72° =108°. 
Then, 

As BD + BC 1175.1 Ar. Co. 6.929925 

Is to BD— BC 121.3 2.083861 

So is tang. ^CD + BDC^ _ ^^^ 10.138739 

^ 2 

To tang. BCD— BDC . go 5, 9.152525 

BDC 45° 55' 

And, 

As sin. BDC 45° 55' Ar. Co. 0.143677 

Is to sin. DBC 72° 9.978206 

So is BC 2.721748 

To CD 697.64 2.843631 

This example might have been solved by finding AD =496.76, 
AC=759.33; whence the angle ADC would be found to be 76° 55', 
and CD =697.64, as before. 



ExAMPiifl 4. (PI. 1, fig. 8.) 
Construction. 

With the given distances construct the triangle ABC. Make ACE 
and CAE respectively equal to 13° 30' and 29° 50'. About the 
triangle AEC describe the circle ACD. Join EB, and produce it to 
meet the circumference in D, which will be the situation of the 
Dbserver. 

Since the angles ADE and ACE are' subtended by the same arc, 
we have ADE=ACE=13° 30'. Also CDE=CAE=29° 50'. 



PRACTICAL QUESTIONS. 19 

Calculation. 

In the triangle ABC, we have the three sides to find the angle 
BAG. Thus, 

BC 262 

AC 404 Ar. Co. 7.393619 

AB 213 Ar. Co. 7.671620 

2)879 

Half sum 439.5 ...... 2.642959 

DifFerence 177.5 ..... 2.249198 

2)19.957396 
Cos. i BAC 17° 48' - - - 9.978698 
BAC 35° 36' 

In the triangle ACE we have the angles and side AC, to find AE. 

As sin. AEC 136° 40' Ar. Co. 0.163523 

Is to sin. ACE 13° 30' - - - 9.368185 

So is AC 404 ... - - 2.606381 

To AE 137.43 .--.....--'. 2.138089 

In the triangle ABE we have the sides AB and AE, and the 
included angle B AE = B AC + C AE = 65° 26'. To find ABE, thus : 
As AB + AE .350.43 ------ Ar. Co. 7.455399 

Is to AB— AE 75.57 '.-.-..,. 1.878349 

So is tang. ^^^_AM^ 5^0 ^y .... 10.192195 

,^ AEB ABE „ . r- -r.rnAn 

To tang. ^ 18° 33' - - - - 9.o25943 

ABE =38° 44' 

In ABD-we have ABD=180°— 38° 44'=141° 16', ADB=13° 30', 
and BAD=38° 44'— 13° 30'=25° 14'. To find AD and DB : 

As sin. ADB 13° 30'. Ar. Co. 0.631815 

Is to sin. BAD 25° 14' - - - 9.629721 

So is AB 213 - - 2.328380 

To BD 389- -.--.- 2.589916 



20 PLANE TRIGONOMETRY. 

And, 

As sin. ADB 13° 30' Ar. Co. 0.631815 

Is to sin. ABD 141° 16' 9.796364 

So is AB 213 - : 2.328380 

To AD 570.9 - - - 2.756559 

Finally, in ADC we have the angle ADC=43° 20', CAD=BAC + 
HAD =60° 50' and the side AC ; to find CD. Thus, 

As sin. ADC 43° 20' Ar. Co. 0.163523 

Is to sin. CAD 60° 50' - 9.941117 

So is AC 404 2.606381 

To CD 514.1 2.711021 



This might have been solved by finding ACB = 28° 14', CE = 
292.87,. whence CBE would have been found to be = 77° 26', BD= 
388.9, DC = 514, and AD = 570.8. 

Example 5. (PI. 1, Fig. 9.) 

Here AD = /BD^— AB' = v/ 1296 = 36. 
And AC = AD + DC = 75 Ans. 

Or, Trigbnometrically; 

As BD .39 - - - Ar. Co. 8.408935 

Is to BA 15 1.176091 

So is radius '- lO.OOOOOO 

To COS. B 67° 23' - 9.585026 

And, 

As radius ' . . . Ar. Co. 0.000000 

Is to sin. B 67° 23' - 9.965248 

So is BD 39 1.591065 

To AD 36 1.556313 

AC - 75 



PRACTICAL QUESTIONS. 21 

Example 6. (PI. 1, fig. 10.) 

The angle ACB = DBC — DAC = 25°. 
Then, 

As sin ACB 25° Ar. Co. 0.374052 

* Is to sin. BAG 26° 30' 9.649527 

So is AB 75 1.875061 

To BC 79.18 1.898640 

To find CD, and BD: 

As radius Ar. Co. 0.000000 

Is to sin. B 51° 30' - - 9.893544 

So is CB - - - - - - - 1.898640 

To CD 61.97 ^[^^ 

And, 

As radius Ar. Co. 0.000000 

Is to COS. B ' 9.794150 

So is CB 1.898640 

To BD 49.29. - - - - - . - . . - . 1.692790 



Example 7. (PI. 1, fig. 11.) 

Here ACB = CAD = 35° and BAC = 55° 
Hence, 

As rad. Ar. Co. 0.000000 

Is to tan. BAC 55° - - 10.154773 

So is AB 143. 2.155336 



To BC 204.2 2.310109 



Example 8. (PI. 1, fig. 12.) 
Construction. 

Make AB=76, the distance from the lower column to the statue's 
base. Erect the perpendiculars AD and BF, making the former= 
50. With D as a centre and distance 86, cross BF in F, which 
will be the head of the statue. 



22 PLANE TRIGONOMETRY. 

Make AI = 64, draw IE parallel to AC, with F as a centre and 
distance 97, cross IE in E, then EC perpendicular to AC, will be the 
higher column. 

Calculation. 

To find FDG and side DG: 

As DF 86 Ar. Co. 8.065502 

Is to FG 76 1.880814 

So is radius 10.000000 

To sin. FDG 62° 5i' - 9.946316 

As radius Ar. Co. 0.000000 

Is to cos. FDG 62° oi' ------- - 9.670300 

So is FD 86 1.934498 ' 

To DG 40.25 1.604798 

To find EFH and FH, we have FE= 97 and EH = GI = GD4 
DI= 54.25. Hence, 

As EF 97 Ar. Co. 8.013228 

Is to EH 54.25 - 1.734400 

So is radius - - - 10.000000 

To sin. EFH 34° 9.747628 

4nd, 

iVs radius - Ar. Co. 0.000000 

is to cos. F 34° • - 9.918574 

So is EF 97 1.986772 

To FH 80.42 1.905346 

To find ED, we have EI = HF + FG= 156.42 and DI=U, 
Hence, 

As IE 156.42 Ar. Co. 7.805707 

Is to ID 14 1.146128 

So is radius - . 10.000000 

To tan. lED 5° 7' - 8.951835 

/ 



PRACTICAL QUESTIONS. 23 

As COS. E 5° 7' - Ar. Co. 0.001734 

Is to rad. - 10.000000 

So is IE 156.42 - - - - 2.194293 



To ED 157.04 - - 2.196027 



Otherwise. 

GD = /FD^— FG» = ^/ 1620 = 40.25. 

GI=GD + DI = 54.25. 

FH = V'FE^— Eff = v/ 6465.9375 = 80.41. 

IE = FH + FG = 156.41. 

DPi = yiE^ + ID='= V 24660.0881 = 157.03. 



24 SURVEYING. [Cha!-. I. 



SURVEYING. 



CHAPTER I. 
DIMENSIONS OF A SURVEY. 



PROBLEM 8. 

Example 2. 

Angle B = 34° + 35° = 69°. 

Example 3. 

Here the first bearing must be reversed, since it is towards th t 
station C. It becomes N. 35° W. Hence C= 180°— (35° +87°) =58^. 

Example 4. 
D = 180° ~ (87° — 58°) = 151°. 

PROBLEM 9. 
Example 2. 



1st side S. 


40J° 


E. 


3d 


N. 29i° 


E. 


N 


54 


E. 




N 54 


E. 




94i 






N. 241 


W. 




180 











N. 85i E. 



Prob. 8.] DIMENSIONSOFASURVEY. 25 





4th 


N. 281° E. 






5th N. 57° W. 






N. 54 E. 








N. 54 E. 






N. 25i W. 








Ill 














180 


- 












S. 69 W. 




/' 


6th 


S. 


470 


W. 










N. 54 


E. 










S. 


7 


E. 








Example 


3. 






1st 


S. 45i° W. 
S. 20i W. 

S. 25 W. 






2d 


N. 50° W. 

S. 20i W. 

N. 70i W. 




3d 


N. 0° W. 
S. 20i W. 

N. 20i W. 






4th 


N. 85° E. 
S. 20J W. 

N. 64i E. 




5th 


S. 47° E. 
S. 20i W 

S. 67i E. 






7th 


N. 51 i° W. 
S. 20i W. 

N. 711 W. 



PROBLEM 10. 

Example 1. 

As sin. bearing 32° 30' - Ar. Co. 0.269784 

Is to radius - ^ - 10.000000 

So is departure 10.96 1.039811 

To distance 20.40 1.309595 

And, 

As -radius - - - - Ar. Co. 0.000000 

Is to cotangent bearing - 10.195813 

So is departure 1.039811 

To difference of latitude 17.20 - 1.235624 



20 SUKVEYING. [Chap. I. 

Example 2. 

As distance 44 Ar. Co. 8.356547 

Is to difference of latitude 34.43 1.536937 

So is radius 10.000000 

To cosine of bearing 38° 31' 9.893484 

And, 

As rad. - - - Ar. Co. 0.000000 

Is to tang, bearing 38° 31' 9.900864 

So is diff. lat. 34.43 - 1.536937 

To departure 27.40 1.437801 

Example 3. 

As cosine of bearing 32° 30' - - - - Ar. Co. 0.073971 

Is to radius 10.000000 

So is diff of lat. 17.21 1.235781 

To distance 20.41 1.309752 

And, 

As radius Ar. Co. 0.000000 

Is to tang, bearing 32° 30' 9.804187 

So is diff latitude 17.21 .-..---- 1.235781 

To departure 10.96 1.039968 

Example 4. 

As diff of lat. 27.92 N. ..... Ar. Co. 8.554085 

Is to departure 5.32 E. - - - - 0.725912 

So is radius 10.000000 

To tang. bear. 10° 47' - - - 9.279997 

And, 

As cosine bearing 10° 47' Ar. Co. 0.007737 

Is to radius 10.000000 

So is diff of lat. 1.445915 

To dist. 28.42 1.453652 

/ 



Prob. 12.] DIMENSIONS OF A SURVEY. 



27 



And, 



Example 5. 

As distance 35.35 Ar. Co. 8.451611 

Is to departure 15.08 - - 1.178401 

So is radius - - 10.000000 

To sin. bearing 25° 15' ....... 9.630012 

As radius .-.------ Ar. Co. 0.000000 

Is to cos. bearing 25° 15' 9.956387 

So is distance 1.548389 

To diff. of lat. 31.97 1.504776 



PROBLEM 12. 



Sta. 


Courses. 


Dist. 


N. 


S. 


E. 


W. 


Cor. 

N. 


Cor. 
E. 


N. 


S. 


E. 


W. 


1 


N. 75 E. 


13.70 


3.54 




13.24 




2 


2 


3.56 




13.26 




2 


N. 20| E. 


10.30 


9.65 




3.61 




1 


1 


9.66 




3.62 




3 


East. 


16.20 






16.20 




2 


2 


.02 




16.22 




4 


S. 33i W. 


35.30 




29.44 




19.49 


5 


5 




29.39 




19.44 


5 


S. 76 W. 


16.00 




3.87 




15.52 


2 


2 




3.85 




15.50 


6 


North. 


9.00 


9.00 








1 


1 


9.01 




.01 




7 


S. 84W. 


11.60 




1.21 




11.54 


2 


2 




1.19 




11.52 


8 


N. 531 w. 


11.60 


6.94 






9.29 


2 


2 


6.96 






9.27 


9 


N. 363 E. 


19.36 


15.51 




11.59 




3 


2 


15.54 




11.61 




10 


N. 224 E. 


14.00 


12.93 




5.36 




2 


2 


12.95 




5.3S 




11 


S. 76| E. 


12.00 




2.75 


11.68 




2 


2 




2.73 


11.70 




12 


S. 15 W. 


10.85 




10.48 




2.81 


2 


1 




10.46 




2.80 


13 


S. 18 W. 


10.62 




10.10 




3.28 


2 


1 




10.08 




3.27 








57.57 


57.85 
57.57 


61.68 


61.93 
61.68 















Error South .28 



.25 Error West- 



28 



SURVEYING. 



[Chap. II. 



CHAPTER 11. 
SUPPLYING OMISSIONS. 



. PROBLEM I. 
Example 2. 



Sta. 


Courses. 


Dist. 


N. 


s. 


E. 


W. 


1 


N. i5r w. 


9.40 


9.05 






2.55 


2 


N. 631 E. 


10,43 


4.61 




9.36 




3 


S. 49 E. 


8.12 




5.33 


6.13 




4 


S. 13i E. 


8.45 




8.22 


1.98 




5 


S. 161 E. 


6.44 




6.17 


1.86 




6 








(6.11) 




(10.64) 


7 


N. 60 W. 


9.72 


4.86 






8.41 


8 


N. 17i W. 


7.65 


7.31 




2.27 










25.83 


25.83 


21.60 


21.60 



Then, 

As diff. lat. 6.11 S. Ar. Co. 9.213959 

Is to depart. 10.64 W. - - 1.026942 

So is radius 10.000000 



To tang, bearing S. 60° 8' W. 



10.240901 



And, As cosine bearing 60° 8' Ar. Co. 0.302785 

Is to radius - 10.000000 

So is diff. lat. 0.786041 



To distance 12.27 1.088826 

Example 3. 



Sta. 


Courses. 


Dist. 


N. 


S. 


E. 


W. 


1 


S. 52° W. 


10.70 




6.59 




8.43 


2 


S. 7i W. 


13.92 




13.80 




1.82 


3 


S. 34i E. 


9.00 




7.44 


5.07 




4 






(27.83) 




(5.18) 












27.83 


10.25 


10.25 



Prob. 1.] 



SUPPLYING OMISSIONS. 



29 



Then, 

As diff. lat. 27.83 Ar. Co. 8.555487 

Is to departure 5.18 ----- 0.714330 

So is radi-us 0.000000 



To tang, bearing N. 10° 33' E. 9.269817 



And, 



As cosine bearing 10° 33' - - - - - Ar. Co. 0.007404 

Is to radius 10.000000 

So is diff. lat. 27.83 1.444513 



To distance 28.31 1.451917 

Example 4. 



Sta. 


Bearing. 


Dist. 


N. 


S. 


E. 


W. 


1 


S. 10° E. 


92.20 




90.80 


16.01 




2 


S. 15 W. 


120.50 




116.39 




31.19 


3 


s. m w. 


205.00 




194.40 




65.05 


4 


8. 7H E. 


68.00 




21.58 


64.49 




5 






















423.17 


80.50 


96.24 
80.50 



15.74 



Then, 

As diff. of latitude 423.17 
Is to departure.,15.74 - - 
So is radius ----- 



Ar. Co. 7.373485 

- - - 1.197005 

- - - 10.000000 



To tans;, bearina; 2° 8' 



8.570490 



And, 



As cosine bearing 2° 8' 
Is to radius - - 'i - 
So is diff lat. 423.17 - 



Ar. Co. 0.000301 

- - - 10.000000 

- - - 2.626515 



To distance 423.46 



2.626816 



30 



SURVEYING. 



[Chap. II. 



PROBLEM 11. 
Example 2. 



Sta. 


Bearing. 


Changed 
Bearing. 


Dist. 


N. 


S. 


E. W. 


1 


S. 40i E. 


N. 85i E." 


31.80 


2.49 




31.70 




2 


N. 54 E. 


North. 




(2.08) 








3 


N. 29i E. 


N. 24f W. 


2.21 


2.01 






.93 


4 


N. 28| E. 


N. 25i W. 


35.35 


31.98 






15.08 


5 


N. 57 W. 


S. 69 W. 






(7.49) 




(1^.51) 


6 


S. 47 W. 


S. 7 E. 


31.30 




31.07 


3.82 












38.56 


38.56 


35.52 


35.52 



As sine changed bearing 69° 
Is to radius ------ 

So is departure 19.51 - - - 

To distance 5th side 20.90 - 



Ar.Co. 0.029848 

- - - 10.000000 

- - - 1.290257 



1.320105 



A.nd, As radius 

Is to cotaug. bearing 69° 
So is departure 19.51 - - 

To diff. latitude 7.49 S. - 



Ar. Co. 0.000000 

- - - 9.584177 

- - - 1.290257 



0.874434 



PROBLEM III. 
Example 2. 



6ta. 


Bearing. 


Changed 
Bearing. 


Dist. 


N. 


S. 


. E. 


W. 


1 


S. 40i E. 


N. 85i E. 


31.80 


2.49 




31.70 




2 


N. 54 E. 


North. 




(2.09) 








3 


N. 29i E. 


N.24|W. 


2.21 


2.01 






.93 


4 


N. E. 




35.35 


(31.97) 






(15.08) 


5 


N. 57 W. 


S. 69 W. 


20.90 




7.49 




19.51 


6 


S. 47 W. 


S. 7 E. 


31.30 




31.07 


3.82 












38.56 


38.56 


35.52 


35.52 



Prob. 4.] 



SUPPLYING OMISSIONS. 



31 



Then, 

As distance 4th side 35.35 - - - Ar. Co. 8.451611 

Is to departure 15.08 1.178401 

So is radius 10.000000 



To sine chang. bearing N. 25° 15' W. 

54 



9.630012 



Bearing 4th side 



N. 28° 45' E. 



And, 



As radius - - - - Ar. Co. 0.000000 

Is to cos. chang. bearing 25° 15' .... 9.956387 
So is distance ---.-- 1.548389 



To diff. latitude 31.97 1.504776 



PROBLEM IV. 

Example 2. (PI. 1, fig. 13.) 



Bearing. 


Dist. 


N. 


S. 


E. 


W. 


FA 


S. E. 


31.80 










AB 


N. 54 E. 


2.08 


1.23 




1.68 




BC 


N. 29i E. 


2.21 


1.92 




1.08 




CD 


N. 281 E. 


35.35 


31.00 




17.00 




DE 


N. 57 W. 


20.90 


11.38 






17.52 


EF 


S. W. 


31.30 










Diff. 


latitude of '. 


EA 


45.53 




19.76 
17.52 


17.52 



Departure of EA 2.24 
Then, As diff. lat. EA 45.53 . - . . Ar. Co. 8.341702 

Is to departure 2.^4 0.350248 

So is radius 10.000000 



To tang, bearing EA 2° 49' 



8.691950 



32 SURVEYING. [Chap. II. 

And, 

As cosine bearing 2° 49' ----- Ar. Co. 0.000525 

Is to radius - 10.000000 

So is diff. lat. - - . - - 1.658298 

To distance EA 45.59 - - - - 1.658823 

To find AEF: 

AF 31.80 

AE 45.59 Ar. Co. 8.341177 

EF 31.30 Ar. Co. 8.504456 
2)108.69 

Half sum 54.34 1.735120 

Difference 22.54 ------ 1.352954 

"""^" 2 )19.933707 

Cos. i AEF 22° 6' - - - 9.966853 

AEF 44° 12' 

Bearing of EA - 2° 49' 

EF S.47°~TW. 

To find EAF and bearing of FA : 

As AF 31.80 Ar. Co. 8.497573 

Is to EF 31.30 1.495544 

So is sin. AEF 44° 12' 9.843336 

To sin. EAF - 43° 20' 9.836453 

Bearing of EA 2° 49' ==" 

AF 40° 31' 



CHAPTER III. 
CONTENT OF LAND. 



PROBLEM I. 

Example 4. 

Here, Area = 176.4 x 176.4 = 31 1 16.96 Sq. Perches, 

= 194 A. 1 R. 36^96 P. 



Proe. 3.] CONTENT OF LAND. 38 



Example 5. 

Here, Area = 52.25 x 38.24 = 1998.04 Sq. Ch, 
= 199 A. 3 R. 8.64 P. 

Example 6. 

Here, Area = 16.54 x 12.37 = 204.5998 Sq. Ck 
= 20 A. 1 R. 33.5968 P. 

Example 7. 

Here, Area = 21.16 x 11.32=239.5312 Sq. Ch. 
= 23 A. 3 R. 32.4992 P. 



PROBLEM 2. 

Example 2 

. 18.37X13.44 246.8928 ,^^..^, ^ ^, 
;re, Area = = = 123.4464 Sq. Ch. 

= 12 A. 1 R. 15.1424 P. 

Example 3. 

49X34 1666 ^ „ 

Here, Area = — ^ — = ~2~ "^ ^^^ ^' 

= 5 A. R. 33 Pe. 



PROBLEM 3. 



•Example 2. (PI. 1, fig. L) 



As radius Ar. Co. 0.000000 

Is to sin. A 47° 30' 9.867631 

AB 15.36 1.186391 

AC 11.46 1.059185 



So is AB X AC 



To double area 129.78 2.113207 

ABC - 64.89 Ch. = 6 A. 1 R. 38. 24 P 



3 



34 SURVEYING. [Chap. Ill, 

Example 3. (PL 1, fig. 14.) 

Here, As radius - Ai\ Co. 0.000000 

Is to sin. A 66° 30' 9.962398 

^ . ,^ ,„(AB 13.84 1.141136 

boisABXAUj^^ 18.23 - 1.260787 

To 2 ABC 231.38 - 2.364321 

ABC - 115.69 Ch. = 11 A. 2 R. 1L04 R 

Example 4. (PI. 1, fig. 15.) 

Here, As radius .--.-..-. Ar. Co. 0.000000 

Is to sin. A 121° 45' ------- - 9.929599 

^ ■ AH &n iAB 19.74 - - - - - 1.295347 

feo IS AJJ, AO j ^^ ^^^^^ ..... 1.239049 

To 2 ABC 291,07 2.463995 

ABC - 145.535 Ch. = 14 A. 2 R. 8.56 P. ^^"""^ 

PROBLEM 4. 

Example 2. (PL 1, fig. 1.) 
Here, Angle C = 180— (A + B) = 43°. Hence, 

A A • n \ radius - - - Ar. Co. 0.000000 

As rad., sm. C | ^.^^ ^ ^^, _ ^^^ ^^^ ^^^^^^^^ 

, , . A • Tj < sin. A 63° - - - . 9.949881 

Is to sm. A, sm. B j ^.^^ ^ ^^, _ _ ^^^^^^^^ 

^ . ,^2 (AB 24.32 - - - - 1.385964 

feo is At. i AB ------ - 1.385964 

To 2 ABC 742.8 2.870868 

ABC - ^L4 Ch. r= 37 A. R. 22.4 P. • """""^ 

Example 3. 
Here, the angle C = 94° 15'. Hence 

A A • r i rad. - - - Ar. Co. 0.000000 

As rad., sm. C j ^.^^ ^ ^^, ^^, ^^^ ^^^ ^^^^^^^^ 

T , . A • -D < sin. A 37° 30' - - - 9.784447 

Is to sm. A. sm. B < . „ „ ^ onL^nrm 

I sin. B 48° 15' - - - 9.872772 

^ • ,r,, (AB 17.36 - - - - 1.239550 

f^o IS AtS j ^g 1.239550 

To 2 ABC 137.25 -...-... 2.137515 
ABC -~68!625 Ch. = 6 A. 3 R. 18 P. 

/ 



Prob. 6.] CONTENT OF LAND. 35 

PROBLEM 5. 

Example 2. 
Here, 10.64 + 12.28 + 9.00 = 3L92 = sum of sides. 

Half sum 15.96 log. 1.203033 

r 5.32 0.725912 

Remainders ) 3.68 0.565848 

( 6.96 0.842609 

2)8.837402 



Area 10)46.63 Ch. ..----- 1.668701 

4.663 = 4 A. 2 R. 26.08 P. 

Example 3. 

Here, 20 + 30 + 40 = 90 = sum of sides. ' 
Half sum 45 - - 1.653218 

/25 1.397940 

Remainders <15 1.176091 

(5 0.698970 

2)4.926214 

10)290.47" 2.463107 

29,047 A. = 29 A. R. 7.52 P. 

PROBLEM 6. 

Example 2, 

Here, 16.10 x ^4^^ = 16.1x5.1 = 82.11 Ch. 

= 8 A. OR. 33.76 P. 

Example 3. 

8 274- 12 43 
Here, 24 X - — ^ — '— = 24 X 10.35 = 248.4 Ch. 

= 24 A. 3R. 14.4 P. 



36 SURVEYING. [Chap. III. 

PROBLEM 7. 
Example 2. 

12.41+8.22 
Here, Area = ~ x 5.15 = 53.12225 Ch. 

= 5 A. 1 R. 9.95G P. 

Example 3. 

11.34+18.46 
Here, Area = ~ x 13.25 = 197.425 Ch. 

= 19 A. 2R. 38.8 P 



Prob. 9.] 



CONTENT OF LAND. 



37 



pq 
O 






C3 






00 

CO 

»o 
It 

T— 1 


1 

1—; 




CO 

CO 


tHO 

(:oco 

CO CO 
CO t^ 

CO 00 

r-l 


U 


o 

CO 
CO 
lO 

T— 1 


o 
o 

CO 






o 
o 

o 
co' 




o 

CO 
CO 

oo' 


ft 

ft 


o 
c4 


o 
o 


r-l 


o 
"*. 

co' 
1—1 


00 

o 

1—1 


05 






^ 


o 

CO 






00 


o 
I— 1 


OS 

CO 

CO 


o 

T— I 


H 




O 

o 


CO 








o 
1—1 


02 








CO 




^ 


rH 

o 

1—1 


;zi 


1— ( 
Oi 


OQ 
l-H 






o 
c<i 




T— 1 

o 
1—1 


u 
o 


o 


o 


rH 


o 


o 


o 


T— I 


a5 

6 


o 


o 


T-i 


o 


o 


o 


I— 1 


^ 


o 

CO 






00 


o 

1—5 


CO 
CO 


o 
I— 1 


w 




o 
o 


CO 








o cr> 
1—1 1— 1 


03 






1—1 


CO 

1-H 




05 


o 
1—1 


^ 


1—1 


00 

I-I 






o 
o4 




1—1 1—1 

d d 
1—1 1— ( 


"S 


CO 




oo 

CO 


Oi 
CO 


CO 


co" 


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d 
CO 




V 

o 

00 


CO 

r-l 

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IC 

CO 


W 

V 
05 

•o 

>— 1 


CO 

o 

CO 
CO 


V 

CO 


V 
05 
lO 
o 

CO 
CO 






■ 1— 1 


C<) 


eo 


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1—1 


o 






o 


o 




o 


o 


o 


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o^ 


Ci 


■^ 


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TtH 


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2. 


-^ 


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c-i 



Ph 



CO 

of 



w 



OQ 



W 



38 




















SURVEYING. 












[Chap. III. 














































o 


o 






Tt< 














CO 


m 








CO 


Oi 


O 


7-< 




TO t^ 


CO 


TO 


TO tH CMTf< 


00 




03 


OS 














CO 


rH 








CO 


lO 


OO 


TO 




rHO 


o 


in 


lO T—< 


TfH 






1—1 














o 


^ 








TO 


(M 


00 


as 




rH (M 


as 


■^ 


^ 00 


C<J 




T-J 














00 


-^ 








CO 


CO 


CO 


00 




<^. ^. 


CO 


CO 


TO TO 


IC 




< 


t>I 














id 


'^' 








rH* 


OO' 


id 


^ 




CO id 


C4 


CD 


CO in 


rH 






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o 


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t^ 


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t^oo 


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C<! 




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1— 1 


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TO 


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in CO 


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TO 


-* 


M^ 


lO 


lO 


lO 


'st* 


TO 










fi 






































Ah' 

lO 










































CM 

iq 
















a 


(O 


1— 1 


05 1 


1 




P-H 


iin 1 iin 


t=^ 








^ 


o 


o 


o 


00 






i 






q 




^ 




rt* 


t- 




T^ 






o 


■^ 




■^ 






1 












TO 




i. 


■^ 


















CD 


F^ 


i- 


TO 1-1 iTO 


00 


■^*" 




-* 










K 










lO 


05 


CO 


CO 


00 


m 


00 


Oi 




TO 








CJ 














CO 




^ 




CO 


CO 


'?} 


CO 




CO 








<i 




CO 


TfH ; 












o" 


~io' 






jTO OS i!> 


oT 










lO 














in 


•^ 






]oo 


O 


Tf 


TO 






^ ■ 




u 

J 
b 


92 


TO 














in 


•^ 








CO 


00 


in 


in 








s 






05 


to" 


r- 


oT 


15" 


to" 






1— 1 


TO in 










■^ 




< 

X 


s: 






irf 


TO 


1?} 
to" 


in 


(7J 






q 


in 
in 












05 
1 




0) 

< 






o 


1 

I— t r-H 


i-H 




rH 


o 


1—1 


1-^ 


1— ( 


rH 


1—1 


O 


1-1 


r-^ 


1— 1 


1-t 


i-t 






O 


1 









































12 









































kl 


t-4 


rH 


1-H 


I— 1 


O 


o 


O 


l-H 


l-H 


1—1 


1— < 


O 


rH 


r-^ 


j-^ 


^' 


rH 










5 
















































"oo" 


To" 




00" 














~~~' 






W 




t 


"^ 


'in 






^ 


1© 


o 




00 




















-rP 




r- 


c^ ^ 








CO 


^' 




^ 




















CO 




CO 


"^j? 


i 




















t^ 


t^ 00 


■^^^ 


c5" 


'^ 


00" 


In" 




i?5" 






0,0 


ts 






H 










lO 


05 


CD 


CO 


00 


in 




00 


05 




TO 






•*c^ 
















CO 




TJH 




CO 


CD 




CQ 


TO 




CO 






oo 


























































iC . 




, 






t^ ICD 








■* 


o" 


oo" 


o" 




rH t^ 


"co 








lO 














in Irt* 








00 


1-1 


^ 


•rr 




Tft CQ 


rH 








03 


CO 














iri 


•* 








CD* 


QD 


in id 




OJOJ 
TO TO 














^ 


cJT 


CD" 


(M 


CD" 


TO" 








CO 


Tn" 








TO 


l^ 








I? 




OD 


TO 


C< 


in 


■^ 








in 


rH 










a> 


<^ 02 

05 










tH 


id 


1-; 


CO 


rt* 


<M" 








in' 


rH 










■*■ 




















































~T 


l> 


a. 





TO 


1—1 


CD 


t^ 


o 


^^ 


"* 


w 


o 


o' 


O 


c* 


¥ 


m 






UJ 


lO 


00 


00 


00 


00 


CO 


C* 


■~D 


00 


in 


in 


T-* 


05 


00 


'^ 


00 


« 








i=l 


l-^ 


id 


in 


00 


<*■ 


"^ 


in 


in 


in 


CO 


in 


TO_ 


t-^ 


GC3 


CO' QD 


CO 










M 


^ 


^ 


x: 


^ 


w 


e4 


e4 


W 


H 




^ 


e4 


K 


^ 


H 


^ 


^ 






S 


CD 


TO 


o 


TO 


CD 




CD 


CO 




1 


o 


00 
CO 


S 


^ 


TO 


o 
in 


t 










CQ 


02 


^ 


^ 


^" 


ai 


^ 


02 


m 




^ 


02 


02 


02 


02' 


^ 










« 


T-H 


(N 


TO 


^ 


lO 


CO 


i> 


00 


a 


o 


rH 


CQ 


TO 


-^ 


in' 


CD 


i> 


























J^ 


f-H 


l-H 


l-H 


r—i 


1— ( 


1—1 


1— 1 










£ 


— 


^_ 







— 


























■■ 











Pros. 9.] 



CONTENT OF LAND. 



X 



























o 


C-5 






















i-O 


USTjH 


OTf< 


o 




C<l 


!M 


C<l 








00 


-^co 


1—1 


o 


o 


c^ 


on 


m 


r-l 


00 


I— 1 








00 


Oi 1-1 


00 


CK) 


05 


CO 


^ 




lO 


O 


o 








CO 


^ CO 


1—1 


lO 


lO 


CO 


Tf 


h 


OO 


^ 


CO 








00 


t^ 


CO 


CO 


^ 


t^ 




o 


Ci 


^ 








-t< 


ai CO 


(M* 


1—1 






1—1 




'^l 


CO 


C<J 








t^ 


t^ CO 


^ 


c^ 


C<) 




cs 




I— 1 


I— 1 

CO 


o 

r— 1 








<^a 




CO 


00 

1—1 


00 

1—1 














CO 


o 


o 




CO 




c! 








1—1 


o 


o 
o 




1— < 
CO 








■< 








CO 


CO 


o 




t^ 














^ 


Ci 


(M 




CO 








^ 








o 


CSI 


o 




CO 














lO 


1—1 


0-D 




(J3 








ft 


^ 


I— 1 


CO 


1—1 


C-. 


O 


t^ 












so 


t- 


1— 1 


b- 


O • 


o 


o 










s 


cc 


CO 


rH 


00 


CO 


c^ 


t^ 












t- 


o 


OS 


CO 




CO 


Tf 










fi 




r-^ 


























'—■ 


1—1 


o 






(M 










t- 


t- 


o 






^ 








^ 








CO 


?o 






00 










<M 


CO 








o 


t^ 


cq 






Gv| 


1—1 








o 


o 


Tt< 








w 


CO 


o 

I— 1 










CO 


00 










«^ 


<N 










-+I 


<rq 






05 


TtH 


C<1 








OO 


■* 








CO 


tH 


C5 


1—1 
1—1 








»o 


00 
















CO 


^ 


lO 




(M 












o 


CO 


t- 




■* 








t< 








CO 
1— 1 


1-1 


CO 

1—1 




^ 










O 


(M 




(M 


tn 


T^^ 


o 






-w 


CO 


rH 




t- 


^ 


05 


CO 










s 


CO 


1—1 




05 


<M 


o 


CO 










<N 


02 




C^ 


<M 


C<1 












sb 


W 


p4 




^ 


^ 


W 


w 










1 


00 


05 
1—1 






1—1 


00 


Ml'* 










m 


m 


OQ 




^ 


^* 


^ 


OQ 










i 


1— 1 


C<1 


CO 


-^ 


to 


CO 


t^ 







































As diff. lat 11.24 Ar. Co. 8.949234 

Is to departure 25.71 1.410102 

So is radius ^ 10.000000 

To tang, bearing 66° 23' - 10.359336 



And, 

Ab cosine bearing 66° 23' Ar. Co. 0.397272 

Is to radius - - - 10.000000 

So is diff. lat. 11.24 1.050766 

To distance 28.06 1.448038 



40 



SURVEYING. 



[Chap. IIL 









OQ 






W i 



W 



o 



-IN 

I— I 



o b- 
CO 





>o 


lOrtH 


oo 


rH 


o 




CO ■* 


00 


05 


o^ 


<x> 


t^ 


CO 


CO 


lO 


CO 


T|H 


^ 


t- 


C<J 


•^ 


<x> 


o 


CO 


T-H 


1—1 


(M 


«5 


00 


oo 




I— 1 









P5 

CO 



^ 



Prob. 9.j C N T E N T F L A N D . 41 

As difF. lat. DF 3.91 Ar. Co. 9.407823 

Is to depart. 19.73 1.295127 

So is radius .....-- 0.000000 

To tang, bearing N. 78° 47' E. - - - - - 10.702950 

As cosine bearing 78° 47' - - - Ar. Co. 0.711036 

Is to radius - 10.000000 

So is diff. latitude 3.91 0.592177 

To distance DF 20.10 r - 1.303213 

FE 19.18 

ED 9.61 --"...■- Ar. Co. 9.017277 

DF 20.10 .-.---." " 8.696804 

2)48.89 

Half sum 24.445 - - - - 1.388190 

Diff. 5.265 ----.--•-- 0.721398 

2)19.823669 

Cos. i FDE 35° 17' 9.911834 

FDE 70° 34' 

Bearing DF N. 78 47 E. 

Bearing DE N. 8° 13' E. 

As FE 19.18 - - -. -8.717151 

Is to DE 9.61 --...-.... 0.982723 

So is sin. FDE 70° 34' -.-.... 9.974525 



To sin. DFE" - - 28° 12' 9.674399 

Bearing FD S. 78 47 W. 

106 59 
180 

Bearing EF S. 73 1 E. 



42 



SURVEYING. 



[Chap. III. 



Example 6. (Fig. 81, Surveying.') 













CO 




00 


1-1 




(M 


00 


lO 1—1 


-* 


(M 


(M T}< 


GOO 


o 


TTl 










05 




^ 


CO 




CO 


lO 


05 Oi 


o 


O 


O 


O'* 


(>J 












o 




'^l 


>o 




00 


CO 


lO CO 


(M 


1— ( 


1— ( 


Tt< 


CO 


2 










CO 




CO 


t^ 




(M 


CO 


CO t- 


CO 


00 


CO 


(M 


CD 


< 










rH 




o 


CO 




lO 


CO 


^ 00 


to 


1^ 


t^ 


T-^ 


Oi 












OS 




CD 


1—1 




o 


Oi 


CO CO 


Oi 


-^ 


^ 


CO 


■^ 












i-H 












CO 


lO CO 


or) 


^ 


^ 




















(M 






(M 




J>-<M 


^ 


c^ 


(M 






























S^t 


































PM 




TJH 


00 


(M 


lO 




O 






(M 






1—1 




(M 


C3 


<» 


to 


CO 


CO 




(M 






t- 






Oi 




S2 


t^ 


o 


o 


•-H 




03 






^ 






CO 






u 


(M 


00 


CD 


lO 




OS 






lO 






t^ 






< 


05 


o 
(N 


o 

I-l 


00 




C3 
T-H 












00 
CO 




■* 


"^ 


t^ 






1— 1 




i-H 






1—1 






CO 




CO 






























■^ 


ri 


(N 


CD 


00 


lO 


rti 


O 


Oi 


CO 


CD 


CD 


CO 










T-H 


o 


'^ 


CD 


CO 


t- 


o 


CD 


lO 


^ 


t- 








a 


o 


lO 




IC 


(M 


CO" 


CD 


t~ 


'^ 


t- 


(M 






ci 




CO 


1—1 




1— ( 


>0 


00 


O 


rH 


OS 


t- 


»o 






a; 


p 














1—1 


tH 












;-i 


































<) 





^ 




CO 

o 

lO 

1—1 












Oi 

o 

00 


00 
Oi 

'^ 

1—1 


c4 


tH 
CD 


CD 
00 

CO 


H 






^ 


Oi 

CD 

1—1 


o 
CO 


CO 

oo' 


CO 

CO 

CO 
1— ( 










CO 
00 

c<5 

CO 


cc 










CD 
CO 




o 


CO 
i-H 




1^ 

00* 


Co" 


CO 
CO 

T-H 

00 


^ 


5 

CO 


00 

r-J 


Oi 

o 
c<i 


Oi 
l-H 




CO 
1—1 
00 
1— 1 






§0 
1-5 






CO 

CO 

00 


1 




CD 
1—1 

>o 

1—1 


o 
1—1 


CO 
00 
00 


o 

CO 


1—1 
o 

CO 


CD 


CO 

00 

c<i 


O 
1-1 


00 






bog 

II 


1 


CO 
00 


p4 

1—1 
1^ 


p4 

rH 


■f4 

O 
00 


p4 


Ml* 

OO 

CO 


Ml* 

o 

C<J 


Ml* 

00 




Ml* 
rH 




tab 


•—1 


rtkM 


o 


•a 
o 


e4 

00 




03 


f4 

o 

CO 


■m 


p4 

CO 


o 
m 




1 


<1 


O 
PC 


p 

O 


§ 








1— ( 


M 
1— 1 




t 





Prob. 9.] CONTENT OF LAND. 

As sine changed bearing LA 71° 45' - - Ar. Co. 0.022414 

Is to radius 10.000000 

So is departure 22.61 1.354301 

To distance LA 23.81 1.876715 

And, 

As radius Ar. Co. 0.000000 

Is to cotang. bearing 9.518184 

So is departure - 1.354301 

To diff. latitude 7.46 - - 0.872485 



43 



Example 7. (Fig. SOj Surveying.) 
To find the third side. 



Sta. 


Bearing. 


Dist. 


N. 


S. 


E. 


W. 


EA 


S. 52 W. 


10.70 




6.59 




8.43 


AB 


S. 7i W. 


13.92 




13.80 




1.82 


BC 


S. 33i E. 


9.00 




7.53 


4.93 












27.92 




10.25 
4.93 



5.32 



As diff. lat. EC 27.92 
Is to depart. 5.32 - 
So is radius - - - 



Ar.Co. 8.554085 

- - - 0.725912 

- - - 10.000000 



To tang, bearing S. 10° 47' W. 

As cosine bearing 10° 47' - ■ 

Is to radius ■ 

So- is diff. lat. 



9.279997 



Ar.Co. 0.007737 

- - - 10.000000 

- - - 1.445915 



To distance 28.42 - 1.453652 



44 



SURVEYING. 



[Chap. III. 



I 1 
I in ■* 



o 


IC ^ 


o 


lO 


t- 


oo 






Oi -^ 


Oi 


^ 


Ci 


(M 


1—1 


lO 


o 


o 




CO 


00 




T-H 












M 







Pk 



Ph 
o 

o 



^ CO 

I I 



o 

CO 






M 



CD 



^ 

^ 
S' 



cq 



Prob. 10.] 



CONTENT OF LAND. 



45 



PROBLEM 10. 

Example 2. 







o 






CD 


§ 


o 






0^ 















o 


in 00 


f^ 


00 t> 


10 


lOTl* 


o -* 


•>* CD 




00 o 


03 






CO T-l 


r-l 


O 03 


Oi 


o> 




CDOJ 


^ 


w^ 


l» 


CO 


fr: 


in rn 




W Tj! 


CD 






in in 


O 


in CO 


(») 








CO 


tH r-t 


Oi 


Ol 




CO >H 




1H 


iH 


t-l 





O O ■<i' CD 



^' 


f4 


e4 


W-f 


-t-f 


^4*1 


en 


CO 


CD 


CO 


lO 


(>i 


^ 


iz; 


^ 


rH 


OJ 


CO 





i© o'olw 


lO 


«^ 




o in 1 in 


cr. 


Ci 






rt -* 




CO 


o 


in 


(rt 




00 ,03 


rH 


01 


iH 


o 






O) 


I-l 


O 


^ 


en 


CO 


< 




tH 


w 


^ 


1-1 




^ 


m 




lO 


o 


lO 


lO 




q 




o 


i> 


CD 


Oi 




3 




CO 


in 


in 


O! 




03 














^ 














■^ 




o 


10 


o 


1-1 




w 




(M 


00 


rH 


00 








^ 


CO 


t- 


CO 




a 














M 














-2 


10 


o 


^ 


10 


en 




ffl 


-in 


m 


OJ 


T)< 


m 




O 


o 


OJ 


ci 


(N 


o 


















o 


o 


10 


in CO 




to 


o 


O! 


o 


1-1 C5 




fi 


o 


•* 


(T) 


in CO 






1 


1-1 1-1 




o 


T-l 


0} 


CO 


■^ 


10 




^ 



















CO 


1 


00 

in 


03 
© 

CO 


!> 


1 

oi 

T-l 

tH 


in 






O 

CO 


© 
in 

CD 




CO 
CD 
CD 


CO 




S 
1 




© 

CD 




C3 
CO 


in 


o 

Tjl 




1 

o 


d 


© 
CO 


© 
1-1 

CO 


CD 
CO 




© 







d 


© 

iH 

CO 


in 00 
""! "=? 
d -ri5 

IH ^ 


ol© 

(>! a 

ai i-i 

1-1 O! 


1 


r-l 


<?} 


CO 


■* 


in 


CO 





© © 


r^ 


© 


(^ 


© O! 


•^ 


■* 






lO 






00 CO 


© 


tH 


■^ 


03 I> 


CO 


'^ 


Tt* 




Ttl 




iH 


<B 








fl - 


;j 


^ 


iS 










i-:; 






m 








^ 


^ 


^ 


cS 


CO 






<u 


-t= _l 


rr! 


■^ 


<1 




CO 


■^ 



!B 5 



CS 

<1 




© 

© 

CO 
CO 
iH 


© 

CD 

ni 


© 

CO 
Oi 

00 


© 

00 

cd 


© 

CD 
rH 


a 

3 
CO 




CO 




o 

CO 




-*^ 

3 

M 




rH 

in 


00 
00 


CO 
CO 




o 


© 
m 
d 


in © 
i> o> 

Oi rH 


d 


s 


© 
© 
d 


in 


© 
© 
d 

rH 


CD 
CO 


1 


iH 


OJ 


CO 


TH 









^ 


^ 


o5 




C^ 


(M 










a? 




CO 


CO 


<! 




J> 


i> 


J 


lO 


lO 




* 


in 


in 




ffi 


© 


© 




O 










© 


■* 


"rn 


o 


•^ 


ft 


o 


CO 




T-{ 


o 


rH 


(N 




tzi 









46 



SUKVEYING. 



[Chap. III. 



Problem 10. (Example 3.) 
Adding 1' to each of the angles, we find the bearings as follows ; 



O) 


THi> 


on 


ao-^ 


(MO 


O) 


coco 




(M 


r-l ■<* 


(^ 


CO CO 




o 


i-H 


CO 


QC-"* 


70 


CO 


?* 


05 


■^ -ctl 


OS 


03 


!>: 


CO 


M 


CO 


CO 














W 











CO I -^jH Tj; I CO 
od w 03 lO ! 



n\< 







-* 


CO 


O 


o 


o 


o o 


O 


O i> i> 






03 


o 


(W 


(<) 


o 


lO 


lO 




O CD CO 






CO 


f- 


(I) 


(»> 


o 


o* 


*-- 




05 t^ CO 


<p 




T-l 






o 


o 


in 


CO 


j-t 


lOOO '^ 


<! 










iH 








CO 


r-i OJ -^ 


a 




l-( 


rH 


m 


nn 


- 


10 


lO 








■^ 


■^ 


CO 


CD 


o 


Oi 


Qi 




m 






















^ 














1 








^ 


CD 


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PROB. li.J CONTENT OF LAND. 47 

PROBLEM 13. 

Example 2. (PI. 2, fig. 5.) 

Here the various angles will be found to be as in the following 
proportions. Then, 

To find log. of GA : 

As sin. FAG 88° 30' - - . . . Ar. Co. 0.000149 

Is to sin. GFA 68° 30' 9.968678 

So is FG 20 ch. - - 1.301030 

To GA - 1.269857 

To find log. GB: 

As sin. FBG 42° x\r. Co. 0.174489 

Is to sin. GFB 24° 9.609318 

So is FG -1.301030 

To GB 1.084832 

To find log. GC: 

As sin. GCF 43° 15' Ar. Co. 0.164193 

Is to sin. GFC 38° 9.789342 

So is FG 1.301030 

To GC 1.254565 

To find log. GD: 

As sin. GDF 44° 30' ..... Ar. Co. 0.154338 

Is to sin. GFD 59° - . ...... 9.933066 

So is GF 1.301030 

To GD - -^- - - - - - 1.388434 

To find log. GE : 

Assin. GEF 35° 30' ..... Ar. Co. 0.236046 

Is.to sin. GFE 103° 30' 9.987832 

So is GF 1.301030 

To GE 1.524908 



48 SUKVEYING. [Chap. III. 

To find 2ABG: 

As radius Ar. Co. 0.000000 

Is to sin. AGB 91° 9.999934 

o • -or- Ai- ^BG 1.084832 

So IS BG, AG j ^^ ^^^3^3^^ 

To 2 ABG 226.268 2.354623 

Tofind2BGC: 

As radius Ar Co. 0.000000 

Is to sin. BGC 15° 15' 9.420007 

^ ' rn rr i ^B - 1.084832 

bo IS GB, GC I ^^ _ 1.254565 

To 2 BGC 57.465 1-759404 . 

To find 2CGD: 

As radius Ar. Co. 0.000000 

Is to sin. CGD 22° 15' 9.578236 

a • nn nn \ ^C 1.254565 

So IS GC, GD I ^j^ 1.388434 

To 2 CGD 166.431 2.221285 

To find 2 DGE : 

As radius ..-..-.. Ar. Co. 0.000000 

Is to sin. DGE 35° 30' 9.763954 

GD 1.388434 



So is GD, GE ^ ^^ 1.524908 

To 2 DGE 475.657 ----.--- 2.677296 

To find 2 EGA : 

As radius - - Ar. Co. 0.000000 

Is to sin. EGA 18° 9.489982 

^n ^.vr CA ^GE 1.524908 

So IS EG, GA I ^^ ^269857 

To 2EGA 192.640 - 2.284747 

2 DGE 475.657 [ — 

2 CGD 166.431 

2 BGC 57.465 

892.193 
2 AGB 226.268 

2)665.925 

ABCDE 332.9625 Ch. = 33 A. 1 R. 7.4 P. 



/ 



Prob. 5.] LAYING OUT AND DIVIDING LAND. 49 

CHAPTER IV. 
LAYING OUT AND DIVIDING LAND. 



PROBLEM 1. 

Example 2. 

Here, 325 Acres = 3250 chains. 
And side = v 3250 = 57 chains. 



PROBLEM 2. 
Example 2. 

Here breadth = ^— ,- = — = 6.25 chains. 

8 chains 8 



PROBLEM 3. 

Example 2. 

Here, 27 A. 3 R. 20 P. = 4460 P. 

And, As 7 : 9 : : 4460 : 5734.2857. 

/5734.2857"== 75.725 = length. 

Also, As 9 : 7 : : 75.725 : 58.897 = breadth. 

PROBLEM 4, 

Example 2. (PI. 2, fig. 6.) 

Here, 114 A. 2 R. 33.4 P. = 1147.0875 chains. 
Also, 71147.0875 + 7.55^ = v/ 1204.09 = 34.7. 
And, 34.7 + 7.55 = 42.25 length. 
34.7—7.55 = 27.15 breadth. 

^ PROBLEM 5. 
Example 3. (PI. 2, fig. 7.) 
Here, 2 Acres = 320 Perches. 



50 SURVEYING. [Chap. IV. 



And, 



. . P . . ( AB 30 P. - Ar. Co. 8.522879 

AS At5, Sin, A j ^.^^ ^ ^j, jg, ^^^ ^^^ 0.023682 

Is to 2 ABC 640 2.806180 

So is radius - - 10.000000 

To AC 22.53 1.352741 



Example 4. (PI. 2, fig. 8.) 

AB 32.26 - Ar. Co. 8.491336 



As AB, sin. A . ^.^^ ^ ^^^ ^^, ^^ ^^_ 0.002801 

Is to ABCD 740 2.869232 

So is radius 10-000000 

To AD 23.09 1.363369 



PROBLEM 6. 

Example 2. (PI. 2, fig. 9.) 

Here, 27 A. IR. 16 P. = 273.5 Ch. 
And, 

As ABC 273.5 - Ar. Co. 7.563043 

Is to BDC 100 2.000000 

So is AB 35.20 1.546543 

To BD 12.87 r - - - 1.109586 



PROBLEM 7. 

Example 2. (PI. 2, fig. 10.) 

Construction. 

Make AB, equal to the greater of the given sides (20). Draw 
BD perpendicular to AB, equal to twice the given area, divided by 
AB (12.39). Through D draw DC parallel to AB. Then if AC 
be made equal to the other given side (16.25), and BC be joined; 
ABC will be the triangle. 

For the Division Line. Make AP = 8.50 the given distance. 
Take AF to AC in the ratio of the part to be cut off to the whole 
area. Join PF, draw BG parallel to it ; then PG will be the division 
line. 



Pros. 9.] LAYING OUT AND DIVIDING LAND. 51 

Demonstration. 
AB : AP : : AG : AF, Therefore, AB. AC : AP. AG : : AC . AG : 
AG.AF : : AC : AF, or AB.AC.sin. A : AP. AG.sia A : : AC : 
AF : : m : n (m being the whole area, and n the part to be cut off.) 
Hence, since AC . AB sin. A = m, AP. AG sin. A = n, and PG is 
.the division line. 

Calculation. 

As ABC 123.9375 Ar. Co. 7.906798 

Is to APG 30 1.477121 

^ ■ Kx> Kn < AB 20 1.301030 

So is AB.AC ] .^ ,„^„ 

\ AC 16.25 1.210853 

To AP. AG - - - - 1.895802 

AP = 8.50 0.929419 

AG = 9.255 0.966383 



PROBLEM 8. 
Example 2. (PI. 2, fig. 11.) 

Here, As BAC 100 ch. Ar. Co. 8.000000 

Is to BDG 45 1.653213 

So is BA^ S ^A 25 1.397940 

(BA 1.397940 

To BD^ - 2)2.449093 

BD = 16.77 1.224546 



PROBLEM 9. 

Example 2. (PI. 2, fig. 12.) 

Here the angles are, A= 71° 45', B = 49° 15', and C = 59°. Hence, 

sin. A 71° 45' Ar. Co. 0.022414 



As sin. A. sin. B , . t» ^r^o-.^, ^-io^m^ 

\ sin. B 49° 15' " " 0.120580 

T , J • r^ < radius 10.000000 

Istorad.sm. C j ^.^^ ^ ^^^ _ _ _ _ ^^3^^^^ 

So is 2 ABC 80 ch. 1.903090 

To AB^ . . ^ 2)1.979150 

AB = 9.763 ......... .989575 



52 SURVEYING. [Chap. IV, 



PROBLEM 10. 

Example 2. (PI. 2, fig. 13.) 

Here the angles A == 99° 30', B = 122°, and P = 41° 30'. 

, , , . . . T. ^ sin. A 99° 30' Ar. Co. 0.005997 

And, As sm. A. sin. B | ^j^. ^ 122° - " " 0.071580 

. ( radius 10.000000 

Is to rad. . sm. P j ^^^^ p ^^o gQ, ... 9.821265 

So is 2 ABCD 50 ch. - - 1.698970 

To fourth term 39.61 - 1.597812 

AB^ - - 36 "="""^ 

CD^ - V 75.61 = 8.695. 

Also, As sin. P 41° 30' Ar. Co. 0.178735 

Is to sin. B 122° -...-.... 9.928420 

So is DC— AB 2.695 0.430559 

To AD 3.449 0.537714 



Example 3. (PI. 3, fig. 1.) 

Here the angles are, A = 90°, B = 73° 30', and P = 16° 30'. 

. ( sin. A 90° - Ar. Co. 0.000000 

Also, As sm. A.sm. B j ^j^^ g ^30 gQ, . » ,, 0.018263 

( radius 10.000000 

Is to rad..sm. P j gj^ p jgo g^. . . . 9.453842 

So is 2 ABCD 160 ch. .---.. . .'2.204120 

To fourth term 47.39 1.675725 

AB^ - 182.25 "^"""^ 

CD = V 134.86 = 11.61. 

And, As sin. P 16° 30' Ar. Co. 0.546658 

Is to sin. B 73° 30' 9.981737 

So is AB— CD 1.89 0.276462 

To AD 6.38 0.804857 



Prob. 14.] LAYING OUT AND DIVIDING LAND. 

PROBLEM 12. 

Example 2. (PI 3, fig. 6.) 

Here, As 2 : 1 : : 60^(100) : EF^ = 50, 

EF = V50 = 7.07. 

And, As BC (10) : EF (7.07) : : AB (15) ; AF = 10.605. 

PROBLEM 13. 

Example 2. (PI. 3, fig. 7.) 

Here the angles are, A = 36^ 30', B = 100° 30', C = 43°, E 
74° 30', and F = 69°. 



As sin. E.sin. F 
Is to sin. C . sin. B 
So is BC^ 



sin. E 74° 30 Ar. Co. 0.016089 

sin. F 69 - - " " 0.029848 

sin. C 43° 9.833783 

sin. B 100° 30' - - - 9.992666 

BC 18.66 1.270912 

BC 1.270912 

To fourth term 259.54 - - 2.414210 

4 

9)1038.16 

EF = v/ 11 5.35 = 10.74. 

As sin. A 36° 30' Ar. Co. 0.225612 

Is to sin. E 74° 30' , 9.983911 

So is EF 10.74- - - 1.031004 

To AF 17.40 1.240527 



PROBLEM 14. 
Example 2. (PI. 3, fig. 8.) 



YABM-JCD^^ 
And, DC— AB (29.4): FE— AB (16.13) : : AD (30) : AF = 16. 16 



Here, EF = \/[ " 2 "^ ' "^ \/ 5796. 18 = 76.13. 



54 



SURVEYING. 



[Chap. IV. 



PROBLEM 16. 

Example 1. (PL 3, fig. 12.) 

The area of this tract may be found to be 858.552 square chains. 
(The latitudes and departures are mostly given in the subsequent oper- 
ations.) 



To find area 


ABODE, and the latitude and departure 


of EA. 




N. 


s. 


E. 


w. 


D. M. D. 


N. Areas. 


S. Areas. 


AB 




9.15 




6.46 


40.86 




373.8690 


BC 


17.21 






17.20 


17.20 


296.0120 




CD 


10.41 




2.89 




2.89 


30.0849 




DE 




3.61 


16.60 




21.38 




77.1818 


EA 




(14.86) 


(6.17) 




42.15 




626.3490 




27.62 


27.62 


23.66 


23.66 




326.0969 
2ABCDEI 
AEI 


1077.3998 
326.0969 




751.3029 

858.552 




2)107.2491 




63.6245 



AE 



14.J 



w. 



5.17 



D. M. D. 



5.17 



N. Areas. 



76.8262 



S. Areas. 



EF 



FA 



.93 



21.49 



21.49 



19.9857 



(15.79) 



(16.32) 26.66 



420.9614 



96.8119 



AEF 



96.8119 



2)324.1495 



162.07475 



As AEF 162.07475 Ar. Oo. 7.790285 

Is to AEI 53.6245 1.729363 

So is lat. EF .93 —1.968483 



To lat. EI .31 —1.488131 



Prob. 16.] LAYING OUT AND DIVIDING LAND. 

As AEF - • - Ar. Co. 7.790285 

Is to AEI 1.729363 

So is depart. EF 21.49 1.332236 

To depart. EI 7.11 - - 0.851884 



55 





N. 


s. 


E. 


w. 


AE 


14.86 






5.17 


EI 


.31 




7.11 




lA 




(15.17) 




(1.94) 



As diff. lat. 15.17 Ar. Co. 8.819014 

Is to depart. 1.94 0.287802 

So is radius - - 10.000000 



To tang, bearing AI 7° 17' 9.106816 



As COS. bearing .------ Ar. Co. 0.003518 

Is to radius 10.000000 

So is diff. lat. 15.17 1.180986 



To dist. AI 15.29 1.184504 



56 MISCELLANEOUS QUESTIONS. [Chap. V. 



- CHAPTER V. 
MISCELLANEOUS QUESTIONS. 



Question 1. 

square yarc 

/ / 2420 \ 
And radius = \/ (^rnTT^) = v/ 770.3081 = 27.75. 



Here 4 Acre = 2420 square yards ; 
2420 \ 



Question 2. 

Construction. 

Make AB (PI. 4, fig. 1) = 40 = one of the given sides, and at A 

320 
draw AL perpendicular to AB and = -rjr = 8 ; through L draw GH 

parallel to AB, and with the centre A and distance = 20 = the other 
given side, describe an arc, cutting GH in D and C ; join AC, BC, 
AD, and BD : then will ABC and ABD answer the conditions of the 
pestion. 

Calculation. 



AE = AF = VAC^ — CE^ = ^400 — 64 = 18.3303 ; and BE = 
AB— AE = 21.6697 ; therefore, BC= v/ BE" + EC^=: V 533.57589809 
= 23.099. Also, BF = AB + AF= 58.3303, and BD = n/BF+FP 

= V 3466.42389809 = 58.876. 

Another Solution. 

Find AE = AF as before. Then, from Geometry, BC' = AB' + 
AC^ - 2 AB.AE = 2000 - 1466.424 = 533.576, and BC = 23.099. 
Also, BD^ = AB^ + AD^ + 2 AB.AF - 2000 + 1466.424 = 3466.424, 
and BF ^ 58.876. 



Chap, v.] MISCELLANEOUS QUESTIONS. 57 

Question 3. 

Here it is evident the number of acres will be inversely as the 
number of square yards in a Perch : 

Therefore, 6^ : 5.5^ :: 110 A. : 92 A. 1 R. 28^ P. Cheshire. 
And T : 5,5^ :: 110 A. : 67 A. 3 R. 25 if P. Irish. 



Question 4. 

28 7 
Here-r7r = — = twice the thickness of the wall, also 840 links = 

554.4 feet=the longer diameter within the walls ; 612 links=:403.92 

, , 7 1670.2 , ,. ., , 

feet — the shorter; 554.4 + -= — - — =: longer diameter outside, and 

7 1218 7fi 

403.92 +^= ^ = shorter. By Frob. 10, Chap. III. the area 

o o 

within the walls = 554.4 x 403.92 x .7854 = 223933.248 X .7854 = 

175877.1729792 ft. = 4 A. R. 6 P. The area to the outside = 

1670.2 1218.76 2035572.952 1598738.9965008 
-^— X X .7854 = X .7854= ^ 

= 177637.6662778 feet. Therefore 177637.666 — 175877.173== 
1760.493 = area the wall stands upon. 



Question 5. 

Here the area of an ellipse whose diameters are 3 and 2 is 4.7124. 
Then, since similar figures are as the squares of their like dimensions, 
we have. As 4.7124 : 160 :: 9 : 305.5768 = square of the longer dia- 
meter; consequently V 305.5768 = 17.481 = longer diameter; and 
3:2:: 17.481 : 11.654 = shorter diameter. 

Question 6. 

Find the area of the triangle whose sides are 9, 8, and 6 ; thus, 

' ^ ' -= 11.5, and 711.5x2.5x3.5x5.5= 7553.4375 =23.525 

square perches. Also, 6 A. 1 R. 12 P. = 1012 P., and 23.525 : 1012 
:: 82 : 2753.1562 = square of the second side; therefore 72753.1562 
= 52.47. Also, 

8:9:: 52.47 : 59.029 = longest side. 

8:6:: 52.47 : 39.353 = shortest side. 



58 



MISCELLANEOUS QUESTIONS. [Chap. V. 



Question 7. (PI. 4, fig. 2.) 
To find ABC : 

AB 27.35 

EC . 31.15 

CA 38.00 

2)96.50 

Half sum 48.25 ' 1.683497 

r 20.90 1.320146 

Remainders j 17.10 1.232996 

( 10.25 -.-.... 1.010724 

2)5.247363 

ABC 420.417 2.623681 

To find ACE: 

AC 38. 

CE 40.10 

EA 22.20 

2)100.30 

Half sum 50.15 - - 1.700271 

( 12.15 -...-.. 1.084576 

Remainders } io.05 1.002166 

(27.95 1.446382 

2)5.233395 

ACE 413.71 ..-....-.. 2.616697 

To find CED: 

CE 40.10 

CD 23.70 

DE 29.25 

2)93.05 

Half sum 46.525 1.667686 

C 6.425 0.807873 

Remainders j 22.825 -.-.--. 1.358410 

(17.275 1.237408 

2)5.071377 

CED 343.311 2.535688 



Chap, v.] MISCELLANEOUS QUESTIONS. 59 

Hence the whole area = 420.418 + 413.71 + 343.308 = 1177.436 
Ch. = 117 A. 2R. 38.976 P. 

Question 8. 
Construction. 
Make AB (PI. 4, fig. 3.) equal to half the given perimeter = 52, 
and bisect it in D ; make DC perpendicular to AB and equal to the 
square root of the given area ; with the centre C and radius equal 
to AD, describe an arc cutting AB in E, complete the rectangle 
AEFG and it will be the one required. The demonstration is evi- 
dent from Geometry. 

Calculation. 



DE = v/CE^— CD^ = ^/ 676— 480 = n/196= 14. 

AE = AD + DE = 26+14 = 40, and EF = EB = 26—14 = 12. 

Question 9. 
Construction. 
Draw any line AC, (PI. 4, fig. 4.) and in it take AE = 20 = given 
difference ; make EF perpendicular to AC = 20 ; join AF and pro- 
duce it to B, making FB = 20 ; then will AB be a side of the square. 

Demonstration. 
Since EA =:EF,the angles FAE and AFE are each equal to half 
a right angle, and AC must be the diagonal of the square. Again 
the triangles CEF and CBF are equal, since they are right angled 
at E and B, and have the hypothenuse and one leg in each equal : 
we have therefore CE = CB = CA— 20. 

Calculation. 



AF =VAE'' + EF2= n/800 = 28.284, and AB = AF + FB = 
48.284 ; hence the area = AB^ = 2331.344656 sq. per. = 14 A. 2 R. 
11.34 P. 

Question 10. 
Construction. 
Let ABCD (PI. 4, fig. 5.) be the given rectangle. In BA and BA 
produced take BH = BC, and AR = I AD. On BR describe the 
semicircle BPR, meeting DA produced in P ; bisect AH in O, and 
with the centre O and radius OP, describe the semicircle EPQ, 
make AG = AQ, complete the rectangle AF, and the thing is done. 



60 MISCELLANEOUS QUESTIONS. [Chap. V. 

Demonstration. 

AF = AEx AG = AEx AQ = AP^ = ABx AR = f ABx AD = 
t AC. Also, BE = BH— HE = BC— AQ = AD— AG = GD. 

Calculation. 
AO = 1 AH = 10 : AP^ = ABx f AD = 6000 ; therefore, OP = 



v/AP^ + OA^ = V6100 = 78.1025; BE = BO— OE = 90—78.1025 
= 11.8975. 

Question 11. 

Construction. 
Let ABD (PI. 4, fig. 6.) be the given circle. Draw the diameter 
AB and radius CD perpendicular to it ; take CF = | AC ; upon BF 
describe a semicircle cutting CD in E : with C as a centre and 
radius CE, describe the circle EGH, and the thing is done. 

Demonstration. 

CE is a mean proportional between CF and CB ; hence CF : CB : : 
CE^ : CB^ : : 4 : 5 ; and since circles are as the squares of their radii, 
we have GEH = j ABD. 

Calculation. 

V5 : v^4 : : AC (75) : EC = Z^^ 

>/5 

^150vA5 ^ 3Q^g ^ 67.082, and DE = DC— EC = 7.918. 
5 

Question 12. 

Construction. 

With the given distances form the triangle ABC, (PI. 4, fig. 7.) 

Upon AB describe the equilateral triangle ABD ; join CD and on it 

describe the equilateral triangle CDE, which will be the one required. 

Demonstration. 

Since BD and BC are by construction two of the given distances ; 
it is only necessary to prove that BE = AC, which is evident from 
the equaUty of the triangles DAC and DBE. 



Chap.V.] miscellaneous questions. 61 



Calculation. 
In the triangle ABC, find the angle BAG, thus, 
BC 10 

AC 7.5 Ar. Co. 9.124939 

AB 12.5 ....... Ar. Co. 8.903090 

2)30. 

15 1.176091 

5 0.698970 

2)19.903090 

Cos. i BAC 26° 34' - . - - . 9.951545 

BAC 53° 8' 

Then in the triangle DAC we have DA and AC, and the angle 
DAC = 113° 8' to find DC, thus. 

As DA + AC 20 - Ar. Co. 8.698970 

Is to DA— AC 5 0.698970 

So is tang. — A±^5^ . 33° 26' .... 9.819684 

^ 2 

To tang. DCA— ADC _ go 33- .... 9.217624 

ACD - 42° 48' 
And, 

As sin. ACD 42° 48' Ar. Co. 0.167848 

Is to sin. DAC 113° 8' - 9.963596 

So is AD 12.5 - - 1.096910 

To DC 16.92 1.228354 

Then in CDE, we have the sides and angles to find the area 

thus. 

As radius Ar. Co. 0.000000 

Is to sin CDE 60° 9.937531 

o • nr^v^rkT? (CD 1.228354 

SoisCDxDE jj^^ ^228354 

To 2 CDE 247.88 2.394239 

123.94 Ch. = 12 A. 1 R. 23.04 P. 



62 



MISCELLANEOUS QUESTIONS. [Chap. V. 



Question 13. 

Construction. 
With the given bearings and distances protract the figure ABCDfg 
PL 4, fig. 8. Join Ag, and with the centres g and A, and distances 
equal to the 4th and 7th sides, describe arcs cutting in G. Join AG 
and gG, and draw DE, EF, and FG respectively parallel and equal 
to gG, Df, and fg. Then will ABCDEFG be the required map. 

Calculation. 
To find the bearing and distance of gA. 





Bearing. 


Dist. 


N. 


S. 


E. 


W. 


AB 


S. 72W. 


24.00 




7.42 




22.83 


EC 


North. 


38.00 


38.00 








CD 


N. 824 E. 


41.00 


5.35 




40.65 




Df 


S. 80 E. 


11.50 




2.00 


11.32 




% 


S. 26 W. 


22.00 




19.77 




9.64 


gA 








(14.16) 




(19.50) 








43.35 


43.35 


51.97 


51.97 



As diff. lat. 14.16 Ar. Co. 8.848937 

Is to departure 19.50 - - - - ■- 1.290035 

So is radius 10.000000 



To tang, bearing gA S. 54° V W. 



10.138972 



As COS. bearing 54° 1' --.--. Ar. Co. 0.230955 

Is to radius - - - - 10.000000 

So is diff. lat. 14.16 1.151063 



To distance gA 24.10 



1.382018 



Chap, v.] MISCELLANEOUS QUESTIONS. 63 

In the triangle AGg we have the sides to find the angles AgG and 

GAg; 

Thus, 

AG 37 

gG 20 Ar. Co. 8.698970 

Ag 24.1 Ar. Co. 8.617982 

2)81.1 

Half sum 40.55 1.607991 

» 

Remainder 3.55 0.550228 

2)19.475171 
Cos. i AgG 56° 52i' - - - - 9.737585 
AgG 113° 45' 

And, 

As AG 37 -..----. - Ar. Co. 8.431798 

Is to gG 20 1.301030 

vSo is sin. AgG 113° 45'.- - - .... 9.961569 

To sin. gAG 29° 39' - - 9.694397 

Applying now the bearing of gA to these angles we will have the 
bearing of gG or DE = S. 59° 44' E, and of GA = S. 83° 40' W 
The area will then be calculated as in the following table, viz. 



64 




MISCELLANEOUS QUESTIONS. 


[Chap. V. 






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Chap, v.] MISCELLANEOUS QUESTIONS. 65 

Question 14. 
Construction. 
Make AB, (PI. 4, fig. 9.) = the given side, and divide it in D, so 
that AD naay be to DB in the ratio of 3 to 2 ; in AB produced, take 
DO a fourth proportional to AD — DB, DB, and AD, and with the 
eentre O and radius OD, describe the semicircle DCE ; make AG 
perpendicular to AB, and equal to tw^ice the area divided by AB = 
6 ; through G draw GF parallel to AB, cutting the circle in C and 
F ; join AC BC, AF and BF ; then will ABC and ABF answer the 
conditions of the question. 

Demonsti^alion. 
Since AD— DB : DB : : AD : DO, we have AD : DB : : AO : DO 
or AO : AD : : DO : DB, therefore, AO : DO : : DO : OB, con- 
sequently (Euclid, F. 6.) AC : BC : : AD : DB : : 3 : 2; and AF : BF 
: AD : DB : : 3 : 2. 

Calculation. 

As 3 + 2 : 15 : : 3 : AD = 9, and DB = 6; also, 9—6 : 6 : : 9 : DO 

= 18, and AO = 9 + 18 = 27, join OC, and OF, and let fall the per- 
pendiculars CL and FP ; then OL = /OC"— CL^ = V324— 36 = 
v288 = 16.9706, and AL = AO— OL = 10.0294; hence AC = 
/AL^ + LC"= n/ 136.58886436 = 11.6871; and as 3 : 2 : : 11.6871: 
BC = 7.7914. Again AP = AO + OP = 43.9706, and AF = 



^/'AP + PF = y 1969.41366436 = 44.3781 ; and as 3 : 2 : : 44.3781 : 
BF = 29.5854 

Question 15. 
Construction. 
Make AB, (PI. 4, fig. 10.) = the given side, and BL = the sum 
of the other sides ; Bisect AB in D, and take DH a third propor- 
tional to 2 AB and'BL ; Draw HE perpendicular to BH and equal 

to — — - = 32. Through E draw EF parallel and equal to BL ; 

join EA and produce it to G, making FG = AB ; draw AC parallel 
to FG, and join BC ; then ABC is the triangle required. 

Demonstration. 
By Construction BL" = 2 AB x DH ; also, in the similar triangles 
EGF and EAC, we have GF (AB) : AC : : EF (BL) : EC (HP). 

5 



66 MISCELLANEOUSQUESTIONS. [Chap. V. 

Hence BL x AC = GF x HP, or 2 BL x AC = 2 GF x HP. Subtracting 
these equals from the precedmg, we have BL^ — 2BLxAC = 
2ABxDH — 2GFxHP = 2ABxDP = (BP + AP)x(BP — AP) = 
BP2_AP2 = BC^— AC^. Hence BL/— 2 BL x AC + AC^ = BC^ and 
BL — AC = BC, or BL ^ BC + AC. 

Cdlculation. 

As 2 AB (100) : BL (85) : : BL (85) : DH = 72.25, and AH = 
DH — AD = 47.25. Now m the right angled triangle AHE, we 
have the sides AH and HE, to find HAE and AE ; thus, 

As AH 47.25 ------- Ar. Co. 8.325598 

Is to HE 32 -------- - - 1.505150 

So is radius 10.000000 

To tang. HAE 34° 6 i' - -. - - - - - 9.830748 

And, 

As COS. HAE 34° 6^ - - - - - Ar. Co. 0.081981 

Is to radius 10.000000 

So is AH - - 1.674402 

To AE 57.07 - - - ^ 1.756383 

Now in the triangle GEF we ha\'c FE, FG, and the angle FEG 
= HAE, to find FGE ; thus, 

As FG 50 - Ar. Co. 8.301030 

Is to FE 85 1.929419 

So is sin. GEF 34° 6 ^ 9.748776 

To sin. FGE 72° 25' ------- - 9.979225 

Finally, in ACE we have AE and the angles to find AC ; thus, 

As sin. ACE 73° 284' - - - - Ar. Co. 0.018319 

Is to sin. AEC 34° 6i' - - ' - - - - - - 9.748776 

So is AE - - - - 1.756383 

To AC 33.3793 1.523478 

And BC = 85 — 33.3793 = 51.6207. 



Chap, v.] MISCELLANEOUS QUESTIONS. 67 

Question 16. 
Construction. 
Make AC (PI. 4, fig. 11.) = 50 = the given diagonal, and on itdescriV-e 

a semicircle ABC ; make AE perpendicular to AC and = ■- = 

24 ; draw EB parallel to AC, cutting the semicircle in B ; join AB, BC, 
and draw CD and DA parallel to them; then will ABCD be the 
rectangle required. 

Demonstration. 

Since ABC is an angle in a semicircie, it is right, and ABCD is <. 
rectangle. Also its area = AC x BF = 1200 perches = 7i acres. 

Calculation. 



FG = v/BG^— BF^' r= V49 = 7 ; AF = AG— GF = 18, and AB 
= VAF' + FB^ = V900 = 30, BC = x/AC^— AB^ = V1600 = 40. 

QuESTIOIV 17. 

Construction. 
Make AB (PI. 4, fig. 12.) = the square root of the given aira, and 
draw CE perpendicular to it : draw BC, making ABC = 30°, make 
AE = AC ; bisect AC in D, and draw EF perpendicular to CE and 
= ED. Complete the parallelogram CEFG, which will be the one 
required. 

Demonstration. 
Since the angle B = 30°, and A = 90°, BC = 2 AC = Cl!i = 4 CD, 
and EF = ED = 3 CD ; therefore FC = VEF^ + EC^ =- 5 CD. Also 
AB^ = BC^ — AC^ = IBC^ = lEC^ = EC x ED ^ EC> EF = 
CEFG. . 

" Calculation. 

Since AB^ = | CE' CE^ = f AB^ = f the given area = 784, and 
CE = 28 ; hence EF ■- | EC = 21. 

Question 18. 

Construction. 
With the given bearings and distances protract the figure ABCD, 
(PI. 4, fig. 13.) and from B draw BP according to the given bearing 



68 MISCELLANEOUS QUESTIONS. [Chap. V. 

and distance of the spring. Produce DA and CB to meet in F, and 
through P draw EH parallel to AD. Bisect AF in G, join EG, and 
draw BM parallel to it, and MN parallel to FE. Make MT perpen- 
dicular to MN, and equal to the square root of the given area. Take 
MU a third proportional to MN and MT ; draw UH parallel to MN, 
cutting AF in I; draw IK perpendicular to AF and equal to EP, and 
with the centre K and distance PH describe an arc cutting AD in 
Q ; draw QPR, and the thing is done. 

Denumstratiov . 

In the similar triangles FGE and FMB, we have FB : FM 
: : FE : FG ; therefore, 15.6, the triangle EFM = BFG ; but EFM= 
1 FMNE, and BFG = h BFA ; hence FMNE = BFA. Again, be- 
cause the triangles EPR, IQS, and PHS are similar, and the homo- 
logous sides EP (IK), IQ, and PH (KQ) form a right angled trian- 
gle, we have from Geometry EPR + IQS=SPH. Add FISPE to 
each, and we have FQR = EFIH. But FBA = EFMN, hence 
BAQR = MNIH = MN.MU =MT2 = the given area. 



Calculation. 

From the bearings of the lines the angles may be found as follow. 
AFB = BEP = 23°, ABF = 84° 30', BAF = 72° 30', EBP - 145° 
30', and EPB = 11° 30'. Then, in the triangle EBP we have all the 
angles and side BP, to find EP and EB : 

Thus, 

As sin. BEP 23° - - Ar. Co.. 0.408122 

Is to sin. EBP 145° 30' - - 9.753128 

So is BP 7.90 0.897627 

To EP 11.452 1.058877 

And, 

As sin. BEP Ar. Co. 0.408122 

Is to sin. BPE 11° 30' 9.299655 

So is BP 0.897627 

To BE 4.031 0.605404 



Chap.V]. miscellaneous questions. 69 

Also, in the triangle ABF, all the angles and side AB are given, 
tofindBFand AF-; 

Thus, 

As sin. AFB 23° Ar. Co. 0.408122 

Is to sin. BAF 72° 30' 9.979420 

So is AB 15.20 1.181844 

To BF 37.101 - - 1.569386 

And, 

As sin. AFB Ar. Co. 0.408122 

Is to sin. ABF 84° 30' 9.997996 

So is AB - . - - 1.181844 

To AF 38.722 1.587962 

And FE = FB— BE = 33.07, and FG = i AF = 19.361 ; 
Also, 

As EF 33.07 .--..--- Ar. Co. 8.480566 

Is to GF 19.361 1.286927 

So is BF 37.101 "1.569386 

To FM 21.721 1.336879 

Now, in the parallelogram MIHN, we have MN = FE = 33.07, 
and IMN = F = 23°, and the area = 100 square chains, to find MI; 
Thus, 

MN 33.07 - Ar. Co. 8.480566 



As MN, sin. IMN ^ ^.^^ j^^ ^3, ^^ ^^^ ^^^^^^2 

Is to radius 10.000000 

So is MIHN 2.000000 

To MI 7.739 - - '- 0. 



Therefore PH = EH — EP = FM+MI — EP = 18.008. Now, 
in the right angled triangle IKQ, we have IK = EP = 11.452, and 
KQ = PH = 18.008, to find IQ; thus, 

KQ+KI = 29.46 log. 1.469233 
KQ— KI = 6.556 0.816639 

2)2.285872 

IQ= 13,898 1.142936 

Hence AQ = FQ — FA = FM+MI + IQ — FA = 4.636. 



70 MISCELLANEOUS QUESTIONS. [Chap. V. 

Question 19. 

Construction. 

With the given bearings and distances, protract the figure ABCD, 
(PL 4, fig. 14;) then, by Prob. 15, Chap. IV. divide ABCD into two 
equal parts by the Hne EF, parallel to CD ; also, by the same pro- 
blem, divide ABCD, and EBAF, each into two equal parts by the 
lines OM and PN, parallel to AD ; join MN, produce it to I, and 
draw OH parallel to IM; join IH, then will EF and IH be the 
division hnes required. 

Demonstration. 

Because PN is parallel to OM, we have IN : NM : : IP : PO 
: : IG : GH, because NG is parallel to HM ; therefore, PG is parallel 
to OH, and consequently to IM. Now since OH is parallel to IM, 
we have IHM = lOM, to each add AIMD, and AIHD = AOMD = 
\ ABCD. In the same manner it may be shown that AIGF = 
i ABEF = i ABCD. 

Calculation. 

Draw EK and IL, each parallel to AD, and MU parallel to AB. 
From the given bearings find the angle A = 78° 30', B = 139° 45', 
C = 78° 45', and D = 63°. By Prob. 15, Chap. IV., find EF and 
AF, thus, 

. • r^ ■ T^ \ sin- C 78° 45' Ar. Co. 0.008426 

Is to sin. A. sin. B j .-"• A 78° 30' - - - .- 9.991193 
i sin. B 139° 45' ... 9.810316 

SoisAB^ j^B 23 ^-3^1^28 

^ AB 1.361728 

To fourth term 383.274 2.583510 

CD" 2161.3201 . 

2)2544.5941 
EF = 1/1272.2970 = 35.67 



Chap, v.] MISCELLANEOUS QUESTIONS. 

And in the triangle ECK, 

As sin. E 38° 15' Ar, Co. 0.208243 

Is to sin. C 78° 45' 9.991574 

So is CD— EF 10.82 - 1.034227 

To FD - 17.14 - - -1.234044 

AD 49.64 

AF 32.50 



71 



Then in the triangle ECK, we have the angles and side EK = 
FD, to find EC, thus, 

As sin. C 78° 45' - - • - - - - Ar. Co. 0.008426 

Is to sin. K 63° - - ' 9.949881 

So is EK 17.14 1.234044 

To CE 15.57 1.192351 

Consequently BE = BC — CE = 14.93. Now by the same pro- 
l>lem find OM, AO, FN and AP, thus, 

. A • -n i ^^"- A - - Ar. Co. 0.008807 

As sin. A. sin. D ^ ^.^^ ^ _ _ ^^_ ^^_ 0.050119 

. „ . ^ ( sin. B 9.810316 

Istosm.B.sm.C j ^.^^ ^ ...... 9.991574 

( BC 30.50 ... - 1.484300 

^^ *^ -^^^ i BC 1.484300 

To fourth term 675.18 2.829416 

AD^ 2464.1296 == 

2)3139.3096 
OM = V 1569.6548 = 39.62 

And, 

As sin. EMD 38° 30' Ar. Co. 0.205850 

Is to sin. D 9.949881 

So is AD— OM 10.02 1.000868 

To AO 14.34 1.156599 



72 MISCELLANEOUS QUESTIONS. [Chap. V. 

And, 

l Sin. F - - - Ar. Co. 0.050119 

. „ . „ ( sin. B 9.810B16 

Is to sm. B.sin. ii J • r, nnn-i^^nA 

( sin. E 9.991574 

. (BE 14.93 .... 1.174060 
So is BE j gj, ^^^^^g(^ 

To Fourth term 161.78 2.208936 

AF^ .... 1056.25 

2)1218.03 
PN ... ^/609.01= 24.68. 

And, 

As sin. HMD 38° 30' . . - - Ar. Oo. 0.205850 

Is to sin. F 9.949881 

So is AF — PN 7.82 ........ 0.893207 

To AP 11.19 1.048938 

Hence OP = AO — AP = 3.15 ; wherefore we have 

OM— PN (14.94) : PN (24.68) : : OP (3.15) : IP = 5.20 ; and AI = 
AP— IP = 5.99. 

In the triangle MUL we have the angle U = A, L = D, and side 
MU = 10 = IP + PO == 8.35, to find ML and UL; thus, 

As sin. ULM 63° Ar. Co. 0.050119 

Is to sin. MUL 78° 30' .... . ■ . 9.991198 
So is MU 8.35 ..." 0.921686 

To ML 9.18 0.962998 

And, 

As sin. MLU 63° Ar. Co. 0.050119 

Is to sin. UML 38° 30' 9.794150 

So is MU 8.35 0.921686 

To UL 5.83 0.765955 



Chap, v.] MISCELLANEOUS QUESTIONS. 73 

Therefore IL = lU+UL = OM+UL = 45.45, and from the simi- 
lar triangles ILM and OMH, we have 

As IL (45.45) : LM (9.18) :: OM (39-62) : MH = 8. 

Now, in the triangle ILH, we have the angle L = D, and sides 
[L and LH = LM4-MH = 17.18, to find the angle LIH; thus, 

As LI + LH G2.63 Ar. Co. 8.203218 

Is to LI — LH 28.27 - - 1.451326 

So is tang. LHI + LjH _ ^go 30' . . . . io.212681 



LIH - ... 22 7 
Bearing IL - - 66 15 

IK - S. 88 22 E. 



To tang. i^5? yi? . 36 23 - - - - 9.867225 

* 2 =^=^= 



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